204 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015 Proposed problems PP24615. 24 Let a, b, c, d, α be a numbers from (0, 1) interval, f : (0, 1) → R a convex and decreasing function. Ten are true the following inequality f � 1 − a 3 � + f � 1 − b 3 � + f � 1 − c 3 � + f � 1 − d 3 � ≥ ≥ f � 1 − a 2 b � + f � 1 − b 2 c � + f � 1 − c 2 a � + f � 1 − d 2 a � . Marius Dr˘ agan PP24616. Find the best k ∈ Z such that �√ n + √ n +1+ √ n +2+ √ n +3 � = �√ 16n + k � for each n ∈ N. Marius Dr˘ agan PP24617. Let x, y, z be an interior numbers. Find all the rest dividing the number x 3 yz + xy 3 z + xyz 3 by 11. Marius Dr˘ agan PP24618. Let a, b, c be the positive numbers such that a = b + c. then is true the following inequality: � 1+ 1 a 2 � a 2 � 1 − 1 a 2 −b 2 −c 2 � ≤ � 1+ 1 b 2 � b 2 � 1+ 1 c 2 � c 2 . Marius Dr˘ agan PP24619. Let a, b, c, d be a positive real numbers such that a ≤ b ≤ c. Then is true the following inequality: � a 2 + 14 � � b 2 + 14 � � c 2 + 14 � ≥ 26 (3a +2b + c + 1) 2 . Marius Dr˘ agan PP24620. Let A, B be two square matrices from C , a, b, c, d a strict natural numbers such that a<b, q the quotiend of dividing of b to a, such that cq �= d and A a B c = A b B d = I n . Then it exist a strictly integer number such that B u = I n . Marius Dr˘ agan 24 Solution should be mailed to editor until 30.12.2018. No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new in sights on past problems.
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204 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
Proposed problems
PP24615. 24 Let a, b, c, d,α be a numbers from (0, 1) interval,f : (0, 1) → R a convex and decreasing function. Ten are true the followinginequality f
�1− a3
�+ f�1− b3
�+ f�1− c3
�+ f�1− d3
�≥
≥ f�1− a2b
�+ f�1− b2c
�+ f�1− c2a
�+ f�1− d2a
�.
Marius Dragan
PP24616. Find the best k ∈ Z such that�√n+
√n+ 1 +
√n+ 2 +
√n+ 3
�=�√
16n+ k�for each n ∈ N.
Marius Dragan
PP24617. Let x, y, z be an interior numbers. Find all the rest dividing thenumber x3yz + xy3z + xyz3 by 11.
Marius Dragan
PP24618. Let a, b, c be the positive numbers such that a = b+ c. then istrue the following inequality:�1 + 1
a2
�a2 �1− 1
a2−b2−c2
�≤�1 + 1
b2
�b2 �1 + 1
c2
�c2.
Marius Dragan
PP24619. Let a, b, c, d be a positive real numbers such that a ≤ b ≤ c. Thenis true the following inequality:�a2 + 14
� �b2 + 14
� �c2 + 14
�≥ 26 (3a+ 2b+ c+ 1)2 .
Marius Dragan
PP24620. Let A,B be two square matrices from C, a, b, c, d a strict naturalnumbers such that a < b, q the quotiend of dividing of b to a, such thatcq �= d and AaBc = AbBd = In. Then it exist a strictly integer number suchthat Bu = In.
Marius Dragan
24Solution should be mailed to editor until 30.12.2018. No problem is ever permanently
closed. The editor is always pleased to consider for publication new solutions or new in sights
on past problems.
Proposed Problems 205
PP24621. Let n be a positive integer. Prov that the equation
22n �
x8 + y8�2n
= z2 + t2 + w2 has the integer solutions.
Marius Dragan
PP24622. We consider the complex number Z such that |z| = 1. Re z ≥ 0,Im z > 0. If we denote x = π
2 arg z and we have��z[x] − 1��+��z[3x] − z[2x]
�� =��z[2x] − z[x]
��+��z[3x] − 1
�� . Then√2 ≥��z[x] − 1
�� ·��z[3x] − z[2x]
�� ·��z[2x] − z[x]
�� ·��z[3x] − 1
�� ≥ 1.
Marius Dragan
PP24623. Let be f : R → (0,+∞) be an increasing concave function. Prove
PP24691. If sinx+ sin y = k (cos y − cosx)2 for x, y ∈ R, then determine allk ∈ R for which sin
�k2x�+ sin
�k2y�= 0.
Mihaly Bencze
PP24692. Determine all a, b, c, d ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for
�abcd�bcda
+�bcda�cdab
+�cdab�dabc
+�dabc�abcd
is a perfect square.
Mihaly Bencze
PP24693. Compute Ak,p =∞�n=1
(−1)n−1 nk
pn .
Mihaly Bencze
PP24694. Determine all xk ∈ R (k = 1, ..., n) for which
[x1]3x2+{x3} + {x2}
3x3+[x4]≥ 4
15[x2]
3x3+{x4} + {x3}3x4+[x5]
≥ 415
−−−−−−−−−−−−[xn]
3x1+{x2} + {x1}3x2+[x3]
≥ 415
, when [·] denote the integer part, and {·}
denote the fractional part.
Mihaly Bencze and Jose Luis Dıaz-Barrero
PP24695. Compute� (cosx−sin2 x) cosxdx
sinx cosx(sinx−cosx)+(1+cos x) cosx .
Mihaly Bencze
Proposed Problems 215
PP24696. If Φ = 1+√5
2 then determine all a, b ∈ R for which the series∞�n=1
1na|sin(nbπΦ)| converges.
Mihaly Bencze
PP24697. Find all polynomials P for whichP (x− 2)P (x− 1)P (x) = P
�x3�for all x ∈ R.
Mihaly Bencze
PP24698. Determine all a, b, c, d ∈ N for which F2an+2b − 2cn− 2d isdivisible by 5 for all n ∈ N , when Fk denote the kth Fibonacci number.
Mihaly Bencze
PP24699. Denote xa, xb, xc the distance of orthocenter H to the sides
BC,CA and AB respectively. Prove that xa + xb + xc ≤ 6
�9s2R3r√
3.
Mihaly Bencze
PP24700. Compute∞�n=1
(p (n− 1) + k)xn−1 if |x| < 1 and p, k ∈ Z.
Mihaly Bencze
PP24701. Find the volume of the region bounded by the surfacex2n+1 + y2n+1 + z2n+1 = (xyz)n when n ∈ N∗.
Mihaly Bencze
PP24702. Determine all k,m ∈ N for which the sum of (m+ k) (kn+ 1)consecutive terms of the Fibonacci sequence is divisible by the(kn+m+ 1)st Lucas number.
Mihaly Bencze
PP24703. In all triangle ABC holds:2(s2−r2−Rr)s2+r2+2Rr
+ 3 3√4Rsr4s ≥ 2.
Mihaly Bencze
216 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24708. If ak ∈ (0, 1] (k = 1, 2, ..., n) , then�
cyclic
1a21+a22
≥ 2[n2 ]+1
[n2 ]+1+n�
k=1ak
.
Mihaly Bencze
PP24709. Let be 1kλ
+ 1(k+1)λ
+ ...+ 1nλ = p
q when k, n,λ ∈ N∗. Determine
all λ ∈ N∗ for which p is odd.
Mihaly Bencze
PP24710. If x, y, z > 0 then
(x+ y + z)�
1x + 1
y + 1z
�≥ 3 + 2 (x+ y + z)
�(xxyyzz)
1x+y+z
xyz .
Mihaly Bencze
PP24711. Computeπ�0
sin2n+1 x sin (kx) dx, when n, k ∈ N.
Mihaly Bencze
Proposed Problems 217
PP24712. Let Z × Z × Z → R be a function which satisfies f (0, 0, 0) = 1and f (n, p, k) + f (n+ 1, p, k) + f (n, p+ 1, k) + f (n, p, k + 1) +f (n+ 1, p+ 1, k) + f (n+ 1, p, k + 1)++f (n, p+ 1, k + 1) + f (n+ 1, p+ 1, k + 1) = 0 for all n, p, k ∈ Z. Prove that|f (n, p, k)| ≥ 1
4 for infinitely many integers n, p, k.
Mihaly Bencze
PP24713. Let Tn =�n+12
�for all n ≥ 1, be the nth triangular number. Show
that for all integers k ≥ 2, the sequence�T kn
�n≥1
does not contain anyinfinite subsequence with all terms in geometric progression.
Mihaly Bencze
PP24714. Solve in N the equation�nk
��n+1k+1
��n+2k+2
�=�3n+33k+3
�.
Mihaly Bencze
PP24715. Find the closed form of�p
�1 + 2
p2+ 4
p4
�−1when the product is
over all primes.
Mihaly Bencze
PP24716. Let ABCD be an equilateral tetrahedron with edge length dinscribed in a sphere. Let P be a point of minor semisphere ABC. LetPA = a, PB = b, PC = c. Is it possible for a, b, c and d to all be distinctpositive integers?
Mihaly Bencze
PP24717. The unsigned Stirling number of the first kind�nk
�is the number
of permutation of [n] that have k cycles. Express (−1)n�k
�n3k
�(−1)k ;
(−1)n�k
�n
3k+1
�(−1)k , (−1)n
�k
�n
3k+2
�(−1)k in function of combinatorial
expressions.
Mihaly Bencze
218 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24718. If λ = limn→∞
n1�0
�12
�nxdx, then compute
limn→∞
n
�λ− n
1�0
�12
�nxdx
�.
Mihaly Bencze
PP24719. Solve in Z the following equation(x+ 3)
�x2 + 3
� �x3 + 3
�= y2 + 15.
Mihaly Bencze
PP24720. Determine all a, b ∈ R for which1�0
(ax+ b)2k+1 dx =1�0
(ax+ b)2p+1 dx when k, p ∈ N.
Mihaly Bencze
PP24721. Compute
limn→∞
n
�2√3π9 − 1−
n�k=1
1(b−a)2n+1
b�a(x− a)n (b− x)n dx
�when 0 < a < b and
n ∈ N.
Mihaly Bencze
PP24722. If z1, z2, z3 ∈ C such that |z1| = |z2| = |z3| = a andz1 + z2 + z3 = b. Determine all a ∈ R and b ∈ C such that
|z − z1|2 + |z − z2|2 + |z − z3|2 = 3�1 + |z|2
�for all z ∈ C.
Mihaly Bencze
PP24723. If ak > 0 (k = 1, 2, ..., n) , thenn�
k=1
ak
ak+nn
n−1≤ n
n+1 .
Mihaly Bencze
PP24724. Solve in R the following systemarctg3x1 · arctg3x2 = arctg3x2 · arctg3x3 = ... = arctg3xn · arctg3x1 = π2
18 .
Mihaly Bencze
Proposed Problems 219
PP24725. Solve in R the following system
�√2 + 2 sinx = 2
54 sin y +
�√2− 2 sin z�√
2 + 2 sin y = 254 sin z +
�√2− 2 sinx�√
2 + 2 sin z = 254 sinx+
�√2− 2 sin y
Ionel Tudor and Mihaly Bencze
PP24726. Solve in N the equation(a+ b)b+c + (b+ c)a+b = (a+ b+ c)a+b+c .
Mihaly Bencze
PP24727. If xk > 1 (k = 1, 2, ..., n) andn�
k=1
xk = n, then
�logx1
�xn−12 + xn−2
2 + ...+ x2 + 1�≥ n(3n−1)
2
Mihaly Bencze
PP24728. Solve in N the equation ab + bc + ca = ba! + cb! + ac!.
Mihaly Bencze
PP24729. If xk > 1√n(k = 1, 2, ..., n) , then
�lognx1x2
�(n− 1)x1 +
x2x1
�≥ 1 + n(n−1)
log3
�
3n�
n�
k=1xk
�n−1� .
Mihaly Bencze
PP24730. If A,B ∈ Mn (Z) and AB = BA, detA = detB = 0, thendetermine all n ∈ N for which det (An +Bn) = xn + yn when x, y ∈ Z.
Mihaly Bencze
PP24731. If x, y, z > 0 and xyz = 1 then determine all λ ∈ R for which� 1+xλn
1+xn ≥ 1 for all n ∈ N.
Mihaly Bencze
PP24732. If ak > 0 (k = 1, 2, ..., n) , then
1n
m�i=1
ai1+ai2+...+ainn√
ai1ai2...a
in
+nm n
√a1a2...an
a1+a2+...+an≥ 2m.
Mihaly Bencze
220 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24733. Let A1A2...An be a convex polygon and S =n�
k=1
ak. Prove that
�n
�S−a2−a3S−2a1
≥ n.
Mihaly Bencze
PP24734. If a, b, c > 0 and ab +
bc +
ca ≤ 11 then
ab+ bc+ ca+ (a+ b+ c)�√
ab+√bc+
√ca�≤ 6 (a+ b+ c)2 .
Mihaly Bencze
PP24735. Solve in (0,+∞) the following
system
(1 +√x)
y= 1 + 2y
√z + x2�
1 +√y�z
= 1 + 2z√x+ y2
(1 +√z)
x= 1 + 2x
√y + z2
.
Mihaly Bencze
PP24736. Solve the following system
x4 + 12y2 + 6 = 6�z2 + 2t
�
y4 + 12z2 + 6 = 6�t2 + 2x
�
z4 + 12t2 + 6 = 6�x2 + 2y
�
t4 + 12x2 + 6 = 6�y2 + 2z
�
Mihaly Bencze and Ionel Tudor
PP24737. In all triangle ABC holds
1).� ma
ha≤
√3(s2−r2−4Rr)
2sr
2).��ma
ha
�2≤ (s2−r2−4Rr)
2
4s2r2
Mihaly Bencze
PP24738. Solve in R the following system:729x+6 ·27y+8 ·27−z = 729y+6 ·27z+8 ·27−x = 729z+6 ·27x+8 ·27−y = 15.
Mihaly Bencze
PP24739. In all triangle ABC holds1).� tgA
2+tg2A≤ R+r
R√3
2).� ctgA
2+ctg2A≤ s
R√3
Mihaly Bencze
Proposed Problems 221
PP24740. In all triangle ABC holds
1).� tg2A
(2+tg2A)(2tg2A+1)≤ s
3R
2).� ctg2A
(2+ctg2A)(2ctg2A+1)≤ R+r
3R
Mihaly Bencze
PP24741. If xk > 0 (k = 1, 2, ..., n) andn�
k=1
1chxk
= 3 then� sh2xk
2+sh4xk≤ 1.
Mihaly Bencze
PP24742. Prove thatn�
k=1
k2(k+1)2
(k2+2)2(k2+2k+3)2≤ n
18(n+1) .
Mihaly Bencze
PP24743. Compute∞�
n1,n2,...,nk=1
1n21+n2
2+...+n2k.
Mihaly Bencze
PP24744. Solve in R the following system:
xy2+2
≤ 1√3(z+1)
yz2+2
≤ 1√3(x+1)
zx2+2
≤ 1√3(y+1)
.
Mihaly Bencze
PP24745. If a, b, c > 0 and 2abc+�
ab ≤ 1 then� ab
(2a2+1)(2b2+1)≤ 1
3 .
Mihaly Bencze
PP24746. Prove that∞�n=0
�n
n2+2
�4< π2
36 .
Mihaly Bencze
PP24747. Solve in N the following system
a�b2 + 3c+ 2
�= x2 + y2
b�c2 + 3a+ 2
�= y2 + z2
c�a2 + 3b+ 2
�= z2 + x2
.
Mihaly Bencze
222 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24748. Let ABC be a triangle. Determine all M ∈ Int (ABC) for whichBMC∡−A∡; AMC∡−B∡; AMB∡− C∡ are the angle of a triangle.
Mihaly Bencze
PP24749. If f : R → R is convex, then�
af�
b1+a+ab
�≥ (a+ b+ c) f (1)
when a, b, c > 0 and abc = 1.
Mihaly Bencze
PP24750. Solve in Z the following system:|x− 2015|+ y2 − 3z = |y − 2015|+ z2 − 3x = |z − 2015|+ x2 − 3y = 4.
Mihaly Bencze
PP24751. Solve in N the equation x2 + y2 = 169n when n ∈ N.
Mihaly Bencze
PP24752. Determine all xk ∈ C (k = 1, 2, ..., n) such that� x1x2...xn−1
1+x1+x1x2+...+x1x2...xn−1= 1.
Mihaly Bencze
PP24753. Compute
� �3x2 − 2y + 5
� �3y2 − 2z + 5
� �3z2 − 2x+ 5
�lnxyzdxdydz.
Mihaly Bencze
PP24754. In all triangle ABC holds12sr ≤� (b+ c)wa ≤ 2s
�3 (s2 + r2 − 8Rr) ≤
√3�
a2.
Mihaly Bencze
PP24755. Determine all f : [0, 1] → (0,+∞) of class C1 for which�1�0
f (x) dx
�n
≥1�0
fn+1 (x) dx when n ∈ N.
Mihaly Bencze
Proposed Problems 223
PP24756. Let f : [0,+∞) → R be a continuous function. Compute
limn→∞
n
�1−
1�0
f (nx) dx
�if lim
x→∞f (x) = 1.
Mihaly Bencze
PP24757. Determine all functions f : [0, 1] → (0,+∞) of class C1 for which
f (e) = 1 and1�0
f2 (x) dx+1�0
dx(f ′(x))2
≤ 2.
Mihaly Bencze
PP24758. Determine all functions f : (0,+∞) → (0,+∞) for which
f ′ (x) f�1y
�= 1
z
f ′ (y) f�1z
�= 1
xf ′ (z) f
�1x
�= 1
y
, for all x, y, z ∈ (0,+∞) .
Mihaly Bencze
PP24759. If ai > 0 (i = 1, 2, ..., n) and k ∈ {1, 2, ..., n} thenn�
i=1ai
n
�
n�
i=1ai
+kn
�
i=1ai
�
cyclic(a1+a2+...+ak)
≥ n+ 1. (A generalization of problem 5.343
Mathematical Reflections).
Mihaly Bencze
PP24760. Prove that∞�n=0
5n�
35n+1−5·35n+4
�
7295n−2435
n−35n+1
= 12 .
Mihaly Bencze
PP24761. If a0 = 0, a1 = 1, a2 = 2, a3 = 6 andan+4 = 2an+3 + an+2 − 2an+1 − an for all n ≥ 0 then denote α (n) =
�ann2
�
when [·] denote the integer part. Compute∞�n=1
1α2(n)
.
Mihaly Bencze
PP24762. In all triangle ABC holds�(|a− c|+ |b− c|)2 ≤ 8
�s2 − 3r2 − 12Rr
�.
Mihaly Bencze
224 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24763. Compute limn→∞
n
�1− (n+ 1)!
�e−
n�k=0
1k!
��.
Mihaly Bencze
PP24764. Solve in R the following system
√2x − 1 +
√3y − 2x +
√2z+1 − 3y = 2x + 1√
2y − 1 +√3z − 2y +
√2x+1 − 3z = 2y + 1√
2z − 1 +√3x − 2z +
√2y+1 − 3x = 2z + 1√
2x − 1 +√3y − 2x +
√2z+1 − 3y = 2x + 1
.
Longaver Lajos and Mihaly Bencze
PP24765. Solve in R the following system(3x + 8 log3 y + 8z) (3y − 8 log3 z + 8x) == (3y + 8 log3 z + 8x) (3z − 8 log3 x+ 8y) == (3z + 8 log3 x+ 8y) (3x − 8 log3 y + 8z) = 121.
Mihaly Bencze and Bela Kovacs
PP24766. In all scalene triangle ABC holds1).� (b−a)rc
PP24771. Determine the left-most digit of the decimal expansion of20172000.
Mihaly Bencze
PP24772. If ak ∈ [−3, 3] (k = 1, 2, ..., n) then determine the maximum of�cyclic
��a21 − a2a3 + 1�� .
Mihaly Bencze
PP24773. In all triangle ABC holds2r�5s2 + r2 + 4Rr
�≤ 3 (R+ r)
�s2 + r2 + 2Rr
�.
Mihaly Bencze
PP24774. In all triangle ABC holds� (ctgA
2 )6
ctg2 A2+ctg2 B
2
≥ s2
2r2.
Mihaly Bencze
PP24775. If xk > 0 (k = 1, 2, ..., n) andn�
k=1
xk = 1, then
n−1�
cyclicx1x2...xn−1
− 1n�
k=1xk
≤ n−2n .
Mihaly Bencze
PP24776. In all triangle ABC holds��3r
s
�3rs ctg
A2 ctg
A2 + 2
�3rs ctg
A2 + 2
�≤ 216.
Mihaly Bencze
PP24777. Prove that exist infinitely many triangles ABC such that
�� 1m
rba
��� 1mrc
b
��� 1mra
c
�≥ 1
34R+r−1
�R2r2
�4R+r.
Mihaly Bencze
226 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24778. Determine all a, b, c ≥ 3 for which(a−1)2(b−1)
(c−2)a−2 + (b−1)2(c−1)
(a−2)b−2 + (c−1)2(a−1)
(b−2)c−2 ≤ ab + bc + ca.
Mihaly Bencze
PP24779. Find all functions f : R → R such that f (0) ∈ Q andf�x+ f2
�y3�+ f3
�z2��
= f6 (x+ y + z) for all x, y, z ∈ R.
Mihaly Bencze
PP24780. Solve in N the equation (2a + b)�2b + a
�= a!b!.
Mihaly Bencze
PP24781. Find all functions f : R → R such that(f ◦ f) (x− y) f
�x2 + xy + y2
�= x3 − yf
�y2�for all x, y ∈ R.
Mihaly Bencze
PP24782. Compute limn→∞
n
2−
1�
0(x2−x−2)
ndx
1�
0
(4x2−2x−2)ndx
.
Mihaly Bencze
PP24783. Compute limn→∞
n
2−
1�
0(2x2−5x−1)
ndx
1�
0
(x2−4x−1)ndx
.
Mihaly Bencze
PP24784. Compute limn→∞
n
2−
n+1�
k=1k! csc
�
π
2k
�
n�
k=1k!scs
�
π
2k
� + 2n
.
Mihaly Bencze
PP24785. In all triangle ABC holds 1s2+r2+2Rr
≥ 18R2 .
Mihaly Bencze
Proposed Problems 227
PP24786. In all triangle ABC holds
� (tgA2 )
n−1(tgB
2 )n−2
tgC2
ctgA2+ctgB
2
≥ 22n−2·3n−1rs for all n ∈ N, n ≥ 2.
Mihaly Bencze
PP24787. In all triangle ABC holds� ctgA
2
1+9tg2 A2tg2 B
2
≥ 12 .
Mihaly Bencze
PP24788. Compute limn→∞
n
�lnπ − 3
2 −n�
k=2
�k2 ln
�1− 1
k2
�+ 1��
Ovidiu Furdul and Mihaly Bencze
PP24789. In all triangle ABC holds� a3
a2+ab+b2≥ 9Rr
s .
Mihaly Bencze
PP24790. Prove thatn�
k=1
�e2 + π2k
�< 2ne
n+π(π2−1)e(π−1) .
Mihaly Bencze
PP24791. Prove thatn�
k=1
�e2 +
�e+ 1
k
�2�< 2ne2n+
en(n+1)2 .
Mihaly Bencze
PP24792. In all triangle ABC holds� ma
a ≥ s2−4Rr−r2
2sR .
Mihaly Bencze
PP24793. In all triangle ABC holds��
tgA2 tg
B2
�4 ≥�rs
�2.
Mihaly Bencze
PP24794. In all triangle ABC holds�� s3+27r3(ctgA
2 )3
s2+9r2(ctgA2 )
2
�2
≥ 9r (4R+ r) .
Mihaly Bencze
228 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24795. If ak > 0 (k = 1, 2, ..., n) andn�
k=1
ak = n, then
n�k=1
�1+a3k1+a2k
�2≥
n�k=1
a2k.
Mihaly Bencze
PP24796. Determine all convex pentagon ABCDE for whichPerimeter (ACE) ≥ 2BD.
Mihaly Bencze
PP24797. Let a, b ∈ Q such that a1k + b
1k + (ab)
1k is also rational.
Determine all k ∈ N∗ for which a1k and b
1k must be rationals.
Mihaly Bencze
PP24798. Let be p ≥ k ≥ 3 prime numbers andA =
�k≤i1<i2<...<ik≤p−1
i1i2...ik. Determine all k ∈ N∗ for which A+ 1 is
divisible by p.
Mihaly Bencze
PP24799. Let ABC be an acute-angled triangle with AC �= BC, letM ∈ Int (ABC) and F the foot of the altitude through C. Furthermore, letP and K be the feet of the perpendiculars dropped from A and Brespectively to the extension of C. The line FM intersects the circumcircleof FPK a second time at L. Determine all M for which ML < MF.
Mihaly Bencze
PP24800. The first n primes are p1 = 2, p2 = 3, ... etc. Set K = pp11 pp22 ...ppnn .Find all positive integers a, b such that K
a(a+1)...(a+b−1) is even, andK
a(a−1)...(a+b−1) has exactly a (a+ 1) ... (a+ b− 1) divisors.
Mihaly Bencze
PP24801. Let (d) a line through point A, and ABC be a triangle. The line(d) intersects the line BC at P . Let Q,R be the symmetrical of P withrespect to the lines AB respectively AC. Determine all (d) for whichBC ⊥ QR.
Mihaly Bencze
Proposed Problems 229
PP24802. Find all non-constant polynomialsP (x) = xn + an−1x
n−1 + ...+ a1x+ a0 with rational coefficients whose rootsare exactly the numbers 1
a0, 1a1, ..., 1
an−1with the same multiplicity.
Mihaly Bencze
PP24803. Find all non-decreasing functions f : R → R for whichf�f�x3�+ y + f
PP24806. Consider a triangle ABC with an inscribed circle with centre Iand radius r. Let CA, CB and CC be circles internal to ABC, tangent to itssides and tangent to the inscribed circle with corresponding radii rA, rB andrC . Show that rλA + rλB + rλC ≥ rλ
3λ−1 for all λ ∈ (−∞, 0] ∪ [1,+∞) . Whathappent if λ ∈ (0, 1)?
Mihaly Bencze
PP24807. If xk > 0 (k = 1, 2, ..., n) then
�cyclic
x21
x2≥ �
cyclic
(x1 − x2)2 +
�n
n�k=1
x2k.
Mihaly Bencze
PP24808. Determine all a, b ∈ N for which for each number(n!)a + 2b; (n!)a + 3b; ...; (n!)a + nb there exist a prime divisor that does notdivide any other number from this set.
Mihaly Bencze
230 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24809. In all triangle ABC holds�
ar2a ≥ 6sr (2R− r) .
Mihaly Bencze
PP24810. Let H0 = 0 and Hn =n�
k=1
1k . Determine all p ∈ N∗ for which
n�k=1
(−1)n−k �nk
�pkHk = pHn −H�
np
�, when [·] denote the integer part.
Mihaly Bencze
PP24811. If ak ∈ R (k = 1, 2, ..., n) then2n�k=1
|sin ak|+2n�k=1
|cos ak|+����sin�
2n�k=1
ak
�����+����cos�
2n�k=1
ak
����� ≤ 2 (2n+ 1) .
Mihaly Bencze
PP24812. In all triangle ABC holds� |cos 3A| ≤ 3
��2 +
√2− s
R
�.
Mihaly Bencze
PP24813. Compute min�ab + bc + ca
�where a, b, c > 0 and abc = 1.
Mihaly Bencze
PP24814. Find all injective functions f : Z → Z that satisfy|f (x)− f (y)|+
��f2 (y)− f2 (z)��+��f3 (z)− f3 (x)
�� ≤≤ |x− y|+
��y2 − z2��+��z3 − x3
�� for all x, y, z ∈ Z.
Mihaly Bencze
PP24815. Denote d (n) the number of different divisors of n, and letan+1 = an + kd (n) where a1 ∈ N∗. Determine all k ∈ N∗ for which doesthere exist an a1 such that k consecutive numbers of the sequence are perfectk powers of natural numbers?
Mihaly Bencze
PP24816. Let be ai, bi, ci ∈ C (i = 1, 2, ..., n) , |ai| = |bi| = |ci| = 1
(i = 1, 2, ..., n) and a = 1n
n�i=1
ai, b =1n
n�i=1
bi, c =1n
n�i=1
ci and
Proposed Problems 231
di = abci + abic+ aibc− 2aibici. Prove thatn�
i=1|di| ≤ n
Mihaly Bencze
PP24817. Compute limn→∞
n
�1
128 −n�
k=2
k2+1k3(k2−1)4
�.
Mihaly Bencze
PP24818. Determine all prime numbers Pk (k = 1, 2, ..., n+ 1) for which
232 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24824. Solve in R the following system:
x212x22 + 2 log2 x3 = x4 + 2x5+2x6
x222x23 + 2 log2 x4 = x5 + 2x6+2x7
−−−−−−−−−−−−−−−x2n2
x21 + 2 log2 x2 = x3 + 2x4+2x5
Mihaly Bencze and Lajos Longaver
PP24825. Determine all f : R → R for which(x+ 1)2 f (x+ 1)− (x+ 2)2 f (−x− 1) = (x+ 1)
�2x2 + 6x+ 5
�for all
x ∈ R.
Mihaly Bencze
PP24826. Determine all n ∈ N∗ for which kn is a perfect k power,pn is aperfect p power and rn is a perfect r power, when k < p < r are positiveintegers.
Mihaly Bencze
PP24827. 1). Let ÷a1, a2, ..., an be an arithmetical progression. Determine
the inverse of the matrice A =
a1 a2 ... an−1 anan a1 ... an−2 an−1
− − − −− −−a2 a3 ... an a1
2). What happens when a1, a2, ..., an is a geometrical progression?
Mihaly Bencze
PP24828. If a = 1+√2
2 then determine all n, k ∈ N for which an + a−k ∈ N.
Mihaly Bencze
PP24829. Solve in R the following system
2 + tgx1 =√3 + 2 [tgx2]
2 + tgx2 =√3 + 2 [tgx3]
−−−−−−−−−−−−2 + tgxn =
√3 + 2 [tgx1]
when [·] denote the integer part.
Mihaly Bencze and Lajos Longaver
Proposed Problems 233
PP24830. Solve in R the following system
�x21 − x22 − 1
� �x22 − x23 + 1
�= 4x1x2�
x22 − x23 − 1� �
x23 − x24 + 1�= 4x2x3
−−−−−−−−−−−−−−−−�x2n − x21 − 1
� �x21 − x22 + 1
�= 4xnx1
.
Mihaly Bencze and Ferenc Olosz
PP24831. If ak > 0 (k = 1, 2, ..., n) andn�
k=1
ak = 1 then
√5 ≤
n�k=1
√2ak + 1 ≤
�n (n+ 2).
Mihaly Bencze
PP24832. Determine all p, n, k ∈ N for which the function
f (x) =
�xn sin
�x−k�− x−p cos
�x−k�if x ∈ (0, 1)
0 if x = 0have primitive
functions.
Mihaly Bencze
PP24833. If xk > 1 (k = 1, 2, ..., n) , then determine all n ∈ N for whichx21
x2−1 +x22
x3−1 + ...+ x2n
x1−1 ≥ 2n+1.
Mihaly Bencze
PP24834. If ak > 0 (k = 1, 2, ..., n) , thenn�
k=1
�a2k + ak + 1
�≥ 3n
n�k=1
ak.
Mihaly Bencze
PP24835. Prove that 2√6n
n+1 ≤n�
k=1
3( 23)
k+2( 3
2)k
k(k+1) ≤ 5nn+1 .
Mihaly Bencze
PP24836. If a, b, x > 0 then axm+n + b ≥ xn (m+ n)��
an
�n � bm
�m� 1n+m
for
all n, n ∈ N∗.
Mihaly Bencze
234 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24837. Prove thatn(n+1)
2 ≤n�
k=1
50k(k+1)2k−45ke(k2+k)k−11e2kk+1
25k(k+1)2k−25ke(k2+k)2−6e2kk+1≤ 11n(n+1)
12 .
Mihaly Bencze
PP24838. Prove that 2n9(n+3) ≤
n�k=1
k+1(k+2)(k+3)(k2+k+1)
≤ n3(n+3) .
Mihaly Bencze
PP24839. Prove that∞�k=1
k2+2k√k2+1
> π2
3 .
Mihaly Bencze
PP24840. Prove thatn�
k=1
k2(k4+2k3+3k2+2k2+2k+2)k2+k+1
≥ n(n+1)(2n+1)3 .
Mihaly Bencze
PP24841. Solve in R the following
system
√x2 + x− 1 +
�y − y2 + 1 = z2 − z + 2�
y2 + y − 1 +√z − z2 + 1 = x2 − x+ 2√
z2 + x− 1 +√x− x2 + 1 = y2 − y + 2
.
Mihaly Bencze
PP24842. Solve in R the following system
x2
3+√
9−y2+ 1
4(3−√9−z2)
= y2
3+√9−z2
+ 14(3−
√9−x2)
= z2
3+√9−x2
+ 1
4�
3−√
9−y2� = 1.
Mihaly Bencze
PP24843. Solve in R the following system
�x2 + 3
�4= 256 (2y − 1)�
y2 + 3�4
= 256 (2z − 1)�z2 + 3
�4= 256 (2x− 1)
.
Mihaly Bencze
Proposed Problems 235
PP24844. Solve in R the following system
√x2 + 1 +
�y3 − 1 = z2 + z + 1�
y2 + 1 +√z3 − 1 = x2 + x+ 1√
z2 + 1 +√x3 − 1 = y2 + y + 1
.
Mihaly Bencze
PP24845. Compute limn→∞
n
�1− 2
n�k=1
2k2−14k4+1
�.
Mihaly Bencze
PP24846. Solve in R the following system
x2 + 13y + 4 = 6 (z + 2)√t
y2 + 13z + 4 = 6 (t+ 2)√x
z2 + 13t+ 4 = 6 (x+ 2)√y
t2 + 13x+ 4 = 6 (y + 2)√z
.
Mihaly Bencze
PP24847. Solve in R the following system�x+ 2 + 2
√y + 1 +
�y + 2− 2
√z + 1 =
�y + 2 + 2
√z + 1+
+�z + 2− 2
√x+ 1 =
�z + 2 + 2
√x+ 1 +
�x+ 2− 2
√y + 1 = 2.
Mihaly Bencze
PP24848. Compute limn→∞
n
�23 −
n�k=2
k3−1k3+1
�.
Mihaly Bencze
PP24849. Solve in R the following system
√x+ 4 +
√y − 4 = 2
�z2 +
√t2 − 16− 6
�
√y + 4 +
√z − 4 = 2
�t2 +
√x2 − 16− 6
�
√z + 4 +
√t− 4 = 2
�x2 +
�y2 − 16− 6
�
√t+ 4 +
√x− 4 = 2
�y2 +
√z2 − 16− 6
�.
Mihaly Bencze
PP24850. Solve in R the following system1
x−√
y2−y− 1
y+√z2−x
= 1y−
√z2−z
− 1z+
√x2−x
= 1z−
√x2−x
− 1
x+√
y2−y=
√3.
Mihaly Bencze
236 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24851. Solve in R the following system
√x− 2 +
√4− y = z2 − 6z + 11√
y − 2 +√4− z = x2 − 6x+ 11√
z − 2 +√4− x = y2 − 6y + 11
.
Mihaly Bencze
PP24852. Solve in R the following system√x2 + 32−2 4
�y2 + 32 =
�y2 + 32−2 4
√z2 + 32 =
√z2 + 32−2 4
√x2 + 32 = 3.
Mihaly Bencze
PP24853. Solve in R the following system√x2 + 9−
�y2 − 7 =
�y2 + 9−
√z2 − 7 =
√z2 + 9−
√x2 − 7 = 2.
Mihaly Bencze
PP24854. Solve in R the following systemx2 + 3y + 4
√z2 + 3t− 6 = y2 + 3z + 4
√t2 + 3x− 6 =
= z2 + 3x+ 4�x2 + 3y − 6 = t2 + 3y + 4
�y2 + 3z − 6 = 28.
Mihaly Bencze
PP24855. Solve in R the following system�3x2 − 2y + 15 +
�3y2 − 2z + 8 =
�3y2 − 2z + 15 +
√3z2 − 2x+ 8 =
=√3z2 − 2x+ 15 +
�3x2 − 2y + 8 = 7.
Mihaly Bencze
PP24856. Solve in R the following system√15− x+
√3− y =
√15− y +
√3− z =
√15− z +
√3− x = 6.
Mihaly Bencze
PP24857. Solve in R the following system3x
y2−4y+1− 2y
z2+z+1= 3y
z2−4z+1− 2z
x2+x+1= 3z
x2−4x+1− 2x
y2+y+1= 8
3 .
Mihaly Bencze
PP24858. Solve in R the following system3x2−1
y + 5y3z2−z−1
= 3y2−1z + 5z
3x2−x−1= 3z2−1
x + 5x3y2−y−1
= 1198 .
Mihaly Bencze
Proposed Problems 237
PP24859. Solve in R the following system
(x+ 1)2 = (y − 3)3
(y + 1)2 = (z − 3)3
(z + 1)2 = (x− 3)3.
Mihaly Bencze
PP24860. Solve in R the following system2x
2y2−5y+3+ 13y
2z2+z+3= 2y
2z2−5z+3+ 13z
2x2+x−3= 2z
2x2−5x+3+ 13x
2y2+y−3= 6.
Mihaly Bencze
PP24861. Solve in R the following system
�3x2 + 5y + 8 = 1 +
�3y2 + 5z + 1�
3y2 + 5z + 8 = 1 +√3z2 + 5x+ 1√
3z2 + 5x+ 8 = 1 +�3x2 + 5y + 1
Mihaly Bencze
PP24862. Solve in R the following system
�x− 1 + 2
√y − 2 = 1 +
�y − 1− 2
√z − 2�
y − 1 + 2√z − 2 = 1 +
�z − 1− 2
√x− 2�
z − 1 + 2√x− 2 = 1 +
�x− 1− 2
√y − 2
.
Mihaly Bencze
PP24863. Solve in R the following system
xy2−y+1
+ 2yz2+z+1
= 1y
z2−z+1+ 2z
x2+x+1= 1
zx2−x+1
+ 2xy2+y+1
= 1
.
Mihaly Bencze
PP24864. Solve in C the following system
2x2 − 8y + 12 =�y2 − 4z − 6
�2
2y2 − 8z + 12 =�z2 − 4x− 6
�2
2z2 − 8x+ 12 =�x2 − 4y − 6
�2.
Mihaly Bencze
PP24865. Solve in Z∗\ {−1} the equation 1x(y2+1)
+ 1(x+1)(y+1) +
1y(x2+1)
= 1.
Mihaly Bencze
238 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
240 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24881. Solve in C the following system
2�x21 + 4 =
�x22 + 1 +
�x23 + 9
2�x22 + 4 =
�x23 + 1 +
�x24 + 9
−−−−−−−−−−−−−2�x2n + 4 =
�x21 + 1 +
�x22 + 9
.
Mihaly Bencze
PP24882. Solve in R the following system
x10 + 20y4 = 32 + 10z6 + 21t5
y10 + 20z4 = 32 + 10t6 + 21x5
z10 + 20t4 = 32 + 10x6 + 21y5
t10 + 20x4 = 32 + 10y6 + 21z5
.
Mihaly Bencze and Ionel Tudor
PP24883. Solve in R the following system
4 cos π2x =
√y + tg π
2z4 cos π
2y =√z + tg π
2x
4 cos π2z =
√x+ tg π
2y
Mihaly Bencze and Ionel Tudor
PP24884. Solve in R the following system
x6 + 3y5 + 3z = 1 + 5t3
y6 + 3z5 + 3t = 1 + 5x3
z6 + 3t5 + 3x = 1 + 5y3
t6 + 3x5 + 3y = 1 + 5z3
.
Mihaly Bencze and Ionel Tudor
PP24885. Let n ∈ N and a, b, c, d ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ; a �= 0 such thatabcd+ cd = 2n − 20. Prove that n ≥ 10 and determine all n ∈ N and abcd forwhich holds the given equality.
Ionel Tudor
PP24886. Solve in R the equation64 cos6 x+ 96 cos5 x− 40 cos3 x+ 6 cosx− 1 = 0.
Ionel Tudor
PP24887. a). If n ∈ N, n ≥ 3 then exist an increasing arithmeticalprogression x1, x2, ..., xn ∈ (−1, 1) such that 2x1 + 2x2 + ...+ 2xn = n
Proposed Problems 241
b). If n = 3 then determine one arithmetical progression which verify a).
Ionel Tudor
PP24888. Prove that the function f : [8,+∞) → R wheref (x) = 4 cos π
2x − tg π2x −√
x is increasing. Solve in N the equation4 cos π
2n =√n+ tg π
2n .
Ionel Tudor
PP24889. If x ∈ R thenx8 + x7 + 3x6 + 5x5 + x4 − 10x3 + 12x2 − 8x+ 16 > 0. Solve in R theequation x10 − 10x6 − 21x5 + 20x4 − 32 = 0.
Ionel Tudor
PP24890. If�1 + 1
x
�x= e
�1−
∞�k=1
bk(x+1)k
�when x > 0 then compute
∞�k=1
11+b2k
.
Mihaly Bencze
PP24891. If�1 + 1
x
�x= e
�1−
∞�k=1
dk
( 1112
+x)k
�, then compute
∞�k=1
11+d2k
.
Mihaly Bencze
PP24892. If
An = 2n−1
8 0 0 04n 4n+ 8 4n 4n
n2 − n n2 − n n2 − 5n+ 8 n2 − 5n−n2 − 3n −n2 − 3n −n2 + n −n2 + n+ 8
PP24927. Let p ∈ {1, 2, ..., n− 1} a given number. Determine all n ∈ N for
which
�n�
k=1
�1
arctg 1√k
�2�= p+ n(n+1)
2 when [·] denote the integer part.
Mihaly Bencze
PP24928. If a, b,λ > 0 thenb�a
√lnxdx ≤ (λ−1)(b−a)
2√λ
+ 12√λln bb
aa .
Mihaly Bencze
PP24929. If x, y, z > 1 then�
log x+y2
�13
�x�≤ 27
�
(x+y)2
8xyz(�
x)3
�logx
x+y2 .
Mihaly Bencze
PP24930. If Fn denote the nth Fibonacci number, then�n�
k=1
1Fk
�2
≥n�
k=1
k+1F 2k.
Mihaly Bencze
PP24931. Prove thatn�
k=1
�arctg 1
kk
�2> n2
n2+1.
Mihaly Bencze
PP24932. If ak ∈ (0, 1) (k = 1, 2, ..., 2n+ 1) then
2n+1�k=1
logak
�1
2n+1
2n+1�k=1
ak
�≥
(2n+1)2n+12n+1�
k=1ak
�
2n+1�
k=1ak
�2n+1 .
Mihaly Bencze
248 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24933. If x > 0 (k = 1, 2, ..., n) andn�
k=1
xk = n, then
n�k=1
xn−1k +
n�k=1
xk + n− 1 ≥ 2�
x1x2...xn−1.
Mihaly Bencze
PP24934. If A,B ∈ Mn (C) such that xBA− yAB = In. Determine allx, y ∈ C for which det (AB −BA) = 0.
Mihaly Bencze
PP24935. If a > 1 and x, t > 0 then ax
x − ax+t
x+t ≤ tx(x+t) .
Mihaly Bencze
PP24936. In all triangle ABC holds�� rb+rc
ra
�λ≥ 3�Rr
�λfor all λ ≥ 1.
Mihaly Bencze
PP24937. Solve in R the following system:
(13− 7x1)�1 + 3
1x2
�= 16 (2 + 7x3)
(13− 7x2)�1 + 3
1x3
�= 16 (2 + 7x4)
−−−−−−−−−−−−−−−−−(13− 7xn)
�1 + 3
1x1
�= 16 (2 + 7x2)
Mihaly Bencze and Gyorgy Szollosy
PP24938. If x0, x1 > 1 and xn+1 = log3 (1 + xn+1 + xn) then computelimn→∞
n (1− xn) .
Mihaly Bencze
PP24939. In all triangle ABC holds� r3a
r2a+2rbrc≥ 4R+r
3 .
Mihaly Bencze
PP24940. In all triangle ABC holds� 1
(tgA2+tgB
2 )(tgB2+tgC
2 )≤ 9
4 .
Mihaly Bencze
Proposed Problems 249
PP24941. Solve in R the following system:
11a − 2b = c2
11b − 2c = a2
11c − 2a = b2
Gheorghe Stoica and Mihaly Bencze
PP24942. If b1 ≥ 2b2 ≥ ... ≥ nbn ≥ 0 then b1 +
�n�
k=1
√bk
�2
≥n�
k=1
(k + 1) bk.
Mihaly Bencze
PP24943. Let ABC be a triangle, A1 ∈ (BC) , B1 ∈ (CA) , C1 ∈ (AB) .The medians AA1, BB1, CC1 intersect second time the circumcircle in pointsD,E and F. Prove that
PP24946. If p ∈ N∗ is given, then compute the integer part ofn�
k=1
p
�p+ 1
k! .
Mihaly Bencze
PP24947. Compute limn→∞
n
�e−1 −
n�p=1
pap
�n−1�
p=1(p+1)ap
where
ap = 1 +n�
k=1
f ′k (0) and fk (x) = kx (1− x) (2− x) ... (k − x) .
Mihaly Bencze
250 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24948. If x0 ∈ [0,π] and (n+ 1)xn =n�
k=0
sinxk then compute
limn→∞
n�√
3−√lnnxn+1
�.
Mihaly Bencze
PP24949. Solve in M3 (R) the equation x3 + x2 + x =
1 1 11 1 11 1 1
.
Mihaly Bencze
PP24950. If a ∈ (0, 1) ∪ (1,+∞) then solve in R the following system
2ax+1 = (a− 1)2 y2 +�a2 − 1
�z + 2a
2ay+1 = (a− 1)2 z2 +�a2 − 1
�x+ 2a
2az+1 = (a− 1)2 x2 +�a2 − 1
�y + 2a
Mihaly Bencze and Gyorgy Szollosy
PP24951. If x, y, z ∈ R and z + y + z = (2k + 1)π, k ∈ Z then determineall a, b, c ∈ R for which
�sin ax sin (by − cz) =
�sin ax (sin by − sin cz)
Mihaly Bencze
PP24952. In all triangle ABC holds�
(wa)b+c ≤
�√3(s2−r2−4Rr)
2s
�4s
.
Mihaly Bencze
PP24953. In all triangle ABC holds�� cos2 B
2+cos2 C
2
cos2 A2
≤�
2�(4R+r)3+s2(2R+r)
2Rs2+ 1�.
Mihaly Bencze
PP24954. In all triangle ABC holds� r2a
(4R+r)ra+3r2b≥ 1
2 .
Mihaly Bencze
PP24955. Determine all a, b ∈ N for which�7a + 51b − 2
� �7b + 51a − 2
�is
divisible by 64.
Mihaly Bencze
Proposed Problems 251
PP24956. Compute∞�n=0
(n!)a((n+2)!)b
((2n+3)!)cif a+ b = c; a, b, c ∈ N.
Mihaly Bencze
PP24957. Prove that�
n�k=0
�nk
�2ak�− 1
n
+
�n�
k=0
�nk
�1ak
�− 1n
=
[n2 ]�k=0
�n2k
��2kk
�ak
(1+n)2k
− 1n
for all a > 0
and n ∈ N∗.
Gyorgy Szollosy
PP24958. In all triangle ABC holds�� sin2 B
2+sin2 C
2
sin2 A2
≥√2 +�
(2R−r)(s2+r2−8Rr)−2Rr2
8Rr2.
Mihaly Bencze
PP24959. Solve in R the equation√60− x+
√65− x+
√156− x = 15.
Gyorgy Szollosy
PP24960. If a, b, c > 0 and a ≥ b+ c then solve the equation√bc+ x+
√ca+ x+
√ab+ x = 3a−b−c
2 .
Mihaly Bencze
PP24961. In all triangle ABC holds�� rb+rc
ra≤ 3�
Rr .
Mihaly Bencze
PP24962. If ak ∈ (0, 1) (k = 1, 2, ..., n− 1) such thatn−1�k=1
ak > 1 then solve
in R the following system:
ax11 + ax2
2 + ...+ axn−1
n−1 = (a1 + a2 + ...+ an)xn
ax21 + ax3
2 + ...+ axnn−1 = (a1 + a2 + ...+ an)
x1
−−−−−−−−−−−−−−−−−−−−axn1 + ax1
2 + ...+ axn−2
n−1 = (a1 + a2 + ...+ an)xn−1
.
Mihaly Bencze
252 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24963. If xk ≥ 1 (k = 1, 2, ..., n) , then
�cyclic
x71+6
x62−3x5
2+5x42−5x3
2+3x22−x2+1
≥ 7n.
Mihaly Bencze
PP24964. Solve in R the following system
arccos x12 + arccos 2x2 = arccos x2
2 + arccos 2x3 = ... =arccos xn
2 + arccos 2x1 =2π3 .
Mihaly Bencze
PP24965. If ak > 0 (k = 1, 2, ..., n) , then�
cyclic
a41+1
2a22−3a2+2≥ 2n n
�n�
k=1
ak.
Mihaly Bencze
PP24966. If ak > 0 (k = 1, 2, ..., n) , then
�cyclic
a61+1
6a42−15a32+20a22−15a2+6≥ n n
�n�
k=1
ak.
Mihaly Bencze
PP24967. If x > 0 and λ ∈ (−∞, 0) ∪ [1,+∞) thenn�
k=1
��x+ k−1
k
� 1λ
�≤�nλ−1 [nx]
� 1λ where [·] denote the integer part.
Mihaly Bencze
PP24968. If ak ≥ 1 (k = 1, 2, ..., n) then� a51+4
a42−2a32+2a23−a3+1≥ 5n.
Mihaly Bencze
PP24969. Let ABC be a triangle. Determine all points M in the plane ofthe triangle such that (
�MA) (
�MA ·MB) ≥�MA ·BC2.
Mihaly Bencze
Proposed Problems 253
PP24970. If a (x, y) = x+y2 , a (x, y) =
√xy and h (x, y) = 2
1x+ 1
y
then solve in
R+ the following system
a (x1, x2) + h (x2, x3) = 2g (x4, x5)a (x2, x3) + h (x3, x4) = 2g (x5, x6)−−−−−−−−−−−−−−−−a (xn, x1) + h (x1, x2) = 2g (x3, x4)
.
Mihaly Bencze
PP24971. Let f : R → R be a convex function. Determine all a, b > 0 anda+ b = 1 such that f (x) + f (y) + f (z) +
�f (ax+ by) ≥ 2
�f (bx+ ay)
for all x, y, z ∈ R.
Mihaly Bencze
PP24972. Let f : R → R be a convex function. Determine all a, b, c > 0and a+ b+ c = 1 such thatf (x)+ f (y)+ f (z)+ f (ax+ by + cz)+ f (ay + bz + cx)+ f (az + bx+ cy) ≥≥ 2f (2bx+ 2cy) + 2f (2by + 2cz) + 2f (2bz + 2cx) for all x, y, z ∈ R.
PP25009. If a, b ∈ C are given, then solve the following system(x+ a) y + b = (y + a) z + b = (z + a) t+ b = (t+ a)x+ b = 0.
Mihaly Bencze
PP25010. If xk > 0 (k = 1, 2, ..., n) then
1).�
cyclic
x21
x2≥
n�k=1
xk +1n
��
cyclic
|x1−x2|√x1
�2
2). nn�
k=1
x2k ≥n�
k=1
x2k +2
n(n−1)
��
1≤i<j≤n|xi − xj |
�2
Mihaly Bencze
PP25011. Let ABCDE be a convex pentagon in which DC = DE andBCD∡ = DEA∡ = 90◦. Determine all points F ∈ (AB) such thatFCE∡ = ADE∡ and FEC∡ = BDC∡.
Mihaly Bencze
PP25012. If x0, x1 > 1 and xn+1 = log3 (1 + xn−1 + xn) for all n ≥ 1 thencompute lim
n→∞n (1− xn) .
Mihaly Bencze
PP25013. Let ABC be an acute triangle, AI,BI, CI intersect thecircumcircle in points A1, B1, C1 and AH,BH,CH intersect the circumcirclein points A2, B2, C2. Prove that
�(AI)a (AH)tgA ≤� (A1I)
a (A2H)tgA .
Mihaly Bencze
PP25014. Compute�
1≤i<j≤n
(2i)!(2j)!i!j!(i+j)! .
Mihaly Bencze
Proposed Problems 259
PP25015. Compute limn→∞
�1≤i<j≤n
�i!j!(i+j)!(2i)!(2j)!
�3.
Mihaly Bencze
PP25016. Let ABC be a triangle, the medians AD,BE and CF intersect
the circumcircle in points M,N and P. Prove that (AM+BN+CP )2
AD+BE+CF ≥ 16√3s
9 .
Mihaly Bencze
PP25017. In all acute triangle ABC holds
�� HAHA1
�ctgA≤�
4s2r2
(s2−(2R+r)2)(s2−4Rr−r2)
� s2−4Rr−r2
2sr
when AH,BH,CH
intersect the circumcircle in points A1, B1, C1.
Mihaly Bencze
PP25018. Prove that n! < exp�n(n+1)
2e
�.
Mihaly Bencze
PP25019. Determine all prime pk (k = 1, 2, ..., n+ 1) for which
1p2n+1
=n�
k=1
1p2k.
Mihaly Bencze
PP25020. If λ = limn→∞
n�k=2
�k ln�2k+32k−3
�− 3�, then compute
limn→∞
n
�λ−
n�k=2
�k ln�2k+32k−3
�− 3��
.
Mihaly Bencze
PP25021. Solve in Z the equation�x+ y
2
� �2y + z
3
� �3z + x
4
�= 91
8 .
Mihaly Bencze
PP25022. Determine all prime p for which exist k, n ∈ N such thatnk + k + 1 and (n+ 1)k + k + 1 is divisible by p.
Mihaly Bencze
260 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25023. If xk > 0 (k = 1, 2, ..., n) andn�
k=1
xk = 1 then
n�k=1
x3k
x2k+xk+1
≥ 1n2+n+1
.
Mihaly Bencze
PP25024. Prove thatn�
k=1
k1k < ne
1e .
Mihaly Bencze
PP25025. Prove thatn�
k=1
eke > n(n+1)
2 .
Mihaly Bencze
PP25026. If a, b, c > 0 then�� a3
a2+1
�2+ 1
16
�� (2a+b+c)3
(2a+b+c)2+16
�2≥ 1
2
�� (a+b)3
(a+b)2+4
�2.
Mihaly Bencze
PP25027. If a, b, c, d are the sides of a convex quadrilateral then�� (a+b+c−d)3
(a+b+c−d)2+1
�2≥�
�(a+b)3
(a+b)2+1
�2.
Mihaly Bencze
PP25028. If a, b, c > 0 and a+ b+ c = 3, then34 +�� a3
a2+1
�2≥ 1
9
�� (2a+b)3
(2a+b)2+9
�2+ 1
9
�� (a+2b)3
(a+2b)2+9
�2.
Mihaly Bencze
PP25029. If a, b, c > 0 then�� a3
a2+1
�2+ 1
9
�� (2a+b)3
(2a+b)2+9
�2≥ 2
9
�� (a+2b)3
(a+2b)2+9
�2.
Mihaly Bencze
PP25030. If a, b, c > 0 and a+ b+ c = 3 then34 +�� a3
a2+1
�2≥ 1
2
�� (a+b)3
(a+b)2+4
�2.
Mihaly Bencze
Proposed Problems 261
PP25031. Solve in N∗ the following equation�1 + 1
x
� �1 + 2
y
� �1 + 3
z
�= 8.
Determine all solutions in Z∗.
Mihaly Bencze and Bela Kovacs
PP25032. If x > 1 thenn�
k=1
1k2
�x+ x2
22+ ...+ xk
k2
�> n
n+1 .
Mihaly Bencze
PP25033. In all triangle ABC holds1). (�
ma)3 ≥�m3
a + 3mambmc
2).��
cos A2
�3 ≥� cos3 A2 + 3s
4R
Mihaly Bencze
PP25034. If a > 0 and a �= 1 thena2n+2−1a2−1
+ a(an−1)a−1 + n+ 1 > an+1−1
a−1 + (n+ 1) a.
Mihaly Bencze
PP25035. Prove that�1 + 6
π
�π �1 + 6
e
�e> 576.
Mihaly Bencze
PP25036. If ak ∈ R (k = 1, 2, ..., n) , n ≥ 3 and a12 < a2
3 < ... < ann+1 then
n�k=1
a1+a2+...+akk2(k+3)ak
< n2(n+1) .
Mihaly Bencze
PP25037. In all triangle ABC holds12
��wa
ma−wa+�
hawa−ha
+�
mama−ha
�≥ wa
ma+ ha
wa+ ma
2ma−haand his
permutations.
Mihaly Bencze
PP25038. Prove thatn�
k=1
(k + 1)��
1 + 1k
�k − 1�< n+ 2n−1
6n + 11n(n+1)12 .
Mihaly Bencze
262 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25039. Prove that 3n−12·3n−1 +
n�k=1
�1 + 1
3k
�3k−1
< 3n+ 32n−18·32n for all n ∈ N∗.
Mihaly Bencze
PP25040. Determine all x, y ∈ R for which
e2√2 ln a ln b ≤ asinx+sin ybcosx+cos y ≤ e2
√ln2 a+ln2 b when a, b > 0.
Mihaly Bencze
PP25041. Prove thatn�
k=1
loga�ak − 1
�loga�ak + 1
�loga�a2k + 1
�< 16n2(n+1)2
27 when a > 1.
Mihaly Bencze
PP25042. If xk > 1 (k = 1, 2, ..., n) , thenn�
k=1
lgn�
n�k=1
xk
�≥
m�k=1
lg xk.
Mihaly Bencze
PP25043. Prove thatn�
k=1
�k√1! + k
√2! + ...+ k
√k!�≤ n(2n2+21n+13)
6 .
Mihaly Bencze
PP25044. Compute λ = limn→∞
1
(2nn )
�n�
k=0
(nk)2
n−k+1
�2
and
limn→∞
n
�λ− 1
(2nn )
�n�
k=0
(nk)n−k+1
�2�.
Mihaly Bencze
PP25045. Prove thatn�
k=1
�3√1 · 2 + 3
√2 · 3 + ...+ 3
�k (k + 1)
�≤ n(n+1)(n+5)
9 .
Mihaly Bencze
PP25046. Compute limn→∞
n
�γ − ln 2−
�n�
k=1
(−1)k−1
k
�nk
�− ln 2k
��when
γ = 0, 92 is the Euler constant.
Mihaly Bencze
Proposed Problems 263
PP25047. Solve in R the following system
5− x =�5− y2
�2
5− y =�5− z2
�2
5− z =�5− x2
�2.
Mihaly Bencze
PP25048. Prove that√1 +
√2 + ...+
√n2 − 1 ≥ 2n(n2−1)
3 .
Mihaly Bencze
PP25049. Compute limn→∞
�n2�arctg(n+1)
n+1 − arctgnn
�− π
2
�.
Mihaly Bencze
PP25050. Compute limn→∞
n
�ln 2−
n�k=1
sinh 1n+k
�.
Mihaly Bencze
PP25051. Compute limn→∞
n
�1−
�
1≤i<j≤n
1ij
�
1≤i<j≤n
1(i+1)(j+1)
�.
Mihaly Bencze
PP25052. Solve in M3 (C)×M3 (C) the following system:�X3 + Y 3 = O3
X−1 + Y −1 = I3.
Mihaly Bencze
PP25053. Compute λ = limn→∞
argzn where zn =�i+ 12
� �i+ 22
�...�i+ n2
�
and limn→∞
n (λ− arg zn) .
Mihaly Bencze
PP25054. Determine all a > 0 for whicha2 ≤ (a− 1)
�a+1a−1
�x+ (a+ 1)
�a−1a+1
�x≤ a3 for all x ∈ [0, 1] .
Mihaly Bencze
264 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25055. Prove thatn�
k=1
�k
�k�k√k ≤ n(15n+17)
2 .
Mihaly Bencze
PP25056. If A,B,C ∈ Mn (z) , n ≥ 3 are nonsingular matrices such that(A∗B∗)∗ = BA, (B∗C∗)∗ = CB, (C∗A∗)∗ = AC, then
detA+ detB + detC <√52 .
Mihaly Bencze
PP25057. Let A ∈ Mn (R) such that TrA = Tr (A∗) = 0. Determine alln ∈ N for which det (An + nIn) det (A
n + In) ≥ 0.
Mihaly Bencze
PP25058. Solve in R the following system
1 + 6x2+y2+z2
= 1x + 1
y + 1z
1 + 6y2+z2+t2
= 1y + 1
z + 1t
1 + 6z2+t2+x2 = 1
z + 1t +
1x
1 + 6t2+x2+y2
= 1t +
1x + 1
y
.
Mihaly Bencze
PP25059. Solve in Z the equation 1 + 6x2+y2+z2
= 1x + 1
y + 1z .
Mihaly Bencze
PP25060. Solve in Z the equation 1 + n(n+1)n�
k=1
x2k
=n�
k=1
1xk.
Mihaly Bencze
PP25061. Prove that�
1≤i<j≤n
3
�(n+1)2
n2ij(i+1)(j+1)≤ (n−1)(n+2)
6 .
Mihaly Bencze
PP25062. In all triangle ABC holds� 1
w2λa
≥ 3�
2s2−r2−4Rr
�λfor all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
Proposed Problems 265
PP25063. In all triangle ABC holds�� a
b+c−a ≥ 4(s2−r2−Rr)s2+r2+2Rr
.
Mihaly Bencze
PP25064. In all triangle ABC holds� ab2
a+b ≤ s2 − r2 − 4Rr.
Mihaly Bencze
PP25065. Determine all n ∈ N for which
�n�
k=1
1|sin k|
�≥ 2016 when [· ]
denote the integer part.
Mihaly Bencze
PP25066. Determine all n ∈ N for which
�n�
k=1
1|cos k|
�≥ 2016 when [· ]
denote the integer part.
Mihaly Bencze
PP25067. Determine all n ∈ N for which
�n�
k=1
tg 1k
�≥ 2016 when [· ] denote
the integer part.
Mihaly Bencze
PP25068. Determine all n ∈ N for which
�n�
k=1
ctg 1k
�≥ 2016 when [· ]
denote the integer part.
Mihaly Bencze
PP25069. Determine all n ∈ N∗ for which ne < (n+1)n+1
nn < (n+ 1) e.
Mihaly Bencze
PP25070. Denote pn the nth prime and xn = p1p1−1 · p2
p2−1 ...pn
pn−1 . Determineall n ∈ N for which xn < 2016.
Mihaly Bencze
266 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25071. If xk ∈ [0, 1] (k = 1, 2, ..., n) thenn�
k=1
�xk (1− xk) +
n�k=1
xk ≤ (1+√10)n
2 .
Mihaly Bencze
PP25072. In all triangle ABC holds� 1√
2ab+√3bc+2ca
≥ 98(s2−r2−4Rr)
.
Mihaly Bencze
PP25073. If ak > 0 (k = 1, 2, ..., n) and a12 < a2
3 < ... < ann+1 then
n�k=1
(a1 + a2 + ...+ ak) >n!(n+3)!an13·22n+1 .
Mihaly Bencze
PP25074. In all triangle ABC holds��a2mbmc
bcma+ bcma
mbmc− a�≥ 4Rsr.
Mihaly Bencze
PP25075. If a, b, c > 0 then��a2+b2
a+b
�3≥ 3abc.
Mihaly Bencze
PP25076. If a, b, c > 1, then��a2
bc + bc− a�3
≥ 3abc.
Mihaly Bencze
PP25077. Solve in Z the equation�x2
yz + yz − x��
y2
zx + zx− y��
z2
xy + xy − z�= xyz.
Mihaly Bencze
PP25078. If xk > 0 (k = 1, 2, ..., n) andn�
k=1
xk = 1 then�
1≤i<j≤nxixj ≤ n−1
2n .
Mihaly Bencze
PP25079. If xj > 0 (j = 1, 2, ..., n) andn�
k=1
xk = 1 compute
max�
1≤i1<...<ik≤nxi1xi2 ...xik .
Mihaly Bencze
Proposed Problems 267
PP25080. Prove thatn�
k=1
k8+k5+1k3+1
≥ n(n+1)2 .
Mihaly Bencze
PP25081. If xk ∈�0, π2�(k = 1, 2, ..., n) , then
�cyclic
11−sinx1 sinx2
≤n�
k=1
1cos2 xk
.
Mihaly Bencze
PP25082. In all triangle ABC holds�� r2a
a
�λ≥ 33λ+1
�r2
2s
�λfor all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP25083. In all triangle ABC holds�� a2b
a3+b
�λ≤ sλ
3λ−1 for all λ ∈ [0, 1] .
Mihaly Bencze
PP25084. In all triangle ABC holds��2a2 + b2 + c2
�≥ 4s2
�s2 + r2 + 2Rr
�2.
Mihaly Bencze
PP25085. Prove that 119 +
π1�0
x2tgxdx ≤ 3 ln 22 − 3π2
32 +
π2�0
tg (sinx) dx.
Mihaly Bencze
PP25086. If n ∈ N∗ and f : [0, 1] → R is a continuous function then∞�k=0
�1�0
x2k(k+1)+nf (x) dx
�2
≤ π2
8
1�0
x2nf2 (x) dx.
Mihaly Bencze
PP25087. In all acute triangle ABC holds�
AtgA ≥ 2sπr3(s2−(2R+r)2)
.
Mihaly Bencze
PP25088. In all triangle ABC holds�
AtgA2 ≥ π(4R+r)
3s .
Mihaly Bencze
268 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25089. In all triangle ABC holds�
Ama�
ma+
�
Aha�
ha+
�
Awa�
wa≤ π.
Mihaly Bencze
PP25090. Determine all continuous functions f : R∗ → R∗ for which2�1
f (x) dx+ e52�1
dxf(x) ≤ e4 + e.
Mihaly Bencze
PP25091. Compute limn→∞
n
�3 ln 3− π√
3− 2
n�k=1
1k(3k−2)
�.
Mihaly Bencze
PP25092. Solve in C the following system 1x + 2
y−1 + 3z−2 + 4
t−3 =
= 1y + 2
z−1 + 3t−2 + 4
x−3 = 1z + 2
t−1 + 3x−2 + 4
y−3 = 1t +
2x−1 + 3
y−2 + 4z−3 = 1.
Mihaly Bencze
PP25093. In all triangle ABC holds�
�
�
tgA2+�
tgC2
�
�
tgB2
1+3tgB2
�
tgA2tgC
2
≥√3.
Mihaly Bencze
PP25094. In all triangle ABC holds 272 ≤� ctgA
2ctgB
2
1−tgA2tgB
2
≤ s2
2r2.
Mihaly Bencze
PP25095. If xk ≥ 0 (k = 1, 2, ..., n) andn�
k=1
xk = n then
n�k=1
exk+2e2xk+exk+1
≥ n(e+2)e2+e+1
.
Mihaly Bencze
PP25096. If a, b ∈ [0, 1] then
a+2a2+a+1
+ b+2b2+b+1
≥ 1 + ab+2(ab)2+ab+1
+(1−a)(1−b)(1−ab)(
√a+2
√b)(
√b+2
√a)
√ab
(a2+a+1)(b2+b+1)((ab)2+ab+1).
Mihaly Bencze
Proposed Problems 269
PP25097. If ak > 0 (k = 1, 2, ..., n) then�
cyclic
a1an−11 +(n−1)a2a3...an
≤
�
n�
k=1ak
�2
n2n�
k=1ak
.
Mihaly Bencze
PP25098. In all triangle ABC holds� 1
sin2 A cos2 A2
≥ 163 .
Mihaly Bencze
PP25099. If ak > 0 (k = 1, 2, ..., n) then� a1an−12 (a2+...+an)
n ≥ nn
(n−1)n(�
a1a2)n−1 .
Mihaly Bencze
PP25100. If ak > 0 (k = 1, 2, ..., n) then��
cyclic
a1a2
�2
≥ 2
�n�
k=1
ak
��n�
k=1
1ak
�− n2.
Mihaly Bencze
PP25101. If 1 ≥ b ≥ 0 and a1 = 0 and 2an+1 = b+ a2n for all n ≥ 1 thencompute lim
n→∞n�1−
√1− b− xn
�.
Mihaly Bencze
PP25102. If ak > 1 (k = 1, 2, ..., n) then�
ax1 loga1a2a3...an =
�n�
k=1
ak
�x
.
Mihaly Bencze
PP25103. If xk > 0 (k = 1, 2, ..., n) and
fa (x1, x2, ..., xn) =(xa
1+xa2+...+xa
k)(xa2+xa
3+...+xak+1)...(x
an+xa
1+...+xak−1)
n�
k=1aak
when
k ∈ {1, 2, ..., n} then fa (x1, x2, ..., xn) ≥ fb (x1, x2, ..., xn) for all a ≥ b > 0.
Mihaly Bencze
PP25104. If ak > 0 (k = 1, 2, ..., n) then� an1
a2a3...an≥
n�k=1
ak.
Mihaly Bencze
270 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25105. If ap > 0 (p = 1, 2, ..., n) and k ∈ {1, 2, ..., n− 1} , then�
cyclic
(a1a2...ak)n+1
(a1a2...an)k ≥� a1...ak.
Mihaly Bencze
PP25106. In all triangle ABC holds2(s2−r2−Rr)s2+r2+2Rr
+ 3
�2Rr
s2+r2+2Rr≥ 2.
Mihaly Bencze
PP25107. In all triangle ABC holds s2+r2
2Rr + 3�
rR ≥ 8.
Mihaly Bencze
PP25108. Solve in Z the equation x3
y+z2+ y3
z+x2 + z3
x+y2= 3a2
1+a when
a ∈ Z ∈ \ {−1} is given.
Mihaly Bencze
PP25109. In all acute triangle ABC holdsR(s2+8Rr+7r2)−2r(s2−r2)
r(s2+2Rr+r2)+ 3
�R(s2−(2R+r)2)
r(s√2+r
√2+2Rr)
≥ 2.
Mihaly Bencze
PP25110. If x, y, z > 1 then�
logxy z +3
��logxy z ≥ 2.
Mihaly Bencze
PP25111. If x, y, z > 0 then� x2+yz
x(y+z) + 2 3
�xyz
(x+y)(y+z)(z+x) ≥ 4.
Mihaly Bencze
PP25112. If xk > 0 (k = 1, 2, ..., n) then
�cyclic
x1x2+...+xn
+n
����n�
k=1xk
�
cyclic(x1+...+xn−1)
≥ n+1n−1 .
Mihaly Bencze
PP25113. If a ≥ 1 then 2�a9 + 2a
�≥ 3a3 + 2
�a6 − 1
�√a6 − 1.
Mihaly Bencze
Proposed Problems 271
PP25114. Solve in R the following system
x+ 12 3√y ≥ 3
2 + (z − 1)√t− 1
y + 12 3√z
≥ 32 + (t− 1)
√x− 1
z + 12 3√t
≥ 32 + (x− 1)
√y − 1
t+ 1
2√
3√x≥ 3
2 + (y − 1)√z − 1
.
Mihaly Bencze
PP25115. Prove thatn�
k=2
√k−1
2k52−3k
32+k
76≤ n−1
2n
Mihaly Bencze
PP25116. If ak > 1 (k = 1, 2, ..., n) then
4n�
k=1
1x2k+
n�k=1
�2x8
k−3x2k+4
x6k−1
�2≥ 4
n3
�n�
k=1
ak
�4
.
Mihaly Bencze
PP25117. Prove that 4n+n�
k=2
�2k3−3k+4k
13
k2−1
�2
≥ 2n(n+1)(2n+1)3 .
Mihaly Bencze
PP25118. If x, y, z > 0 then(x+ y) (y + z) (z + x) ≥ 8xyz + 1
3 (�√
x |y − z|)2 .
Mihaly Bencze
PP25119. If x, y, z > 0 then�� x
y+z
�2+ 14xyz
(x+y)(y+z)(z+x) ≥ 4.
Mihaly Bencze
PP25120. If ak > 1 (k = 1, 2, ..., n) and λ ≥ 1 thenn�
k=1
ak +1
n�
k=1
aλk
≥n�
k=1
1aλ−1k
.
Mihaly Bencze
272 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25121. If a ∈�0, π2�then
3
(1+sin2 a)(1+cos2 a)+ 2 sin (sin a+ cos a) cos (sin a− cos a) < 1
sin2 a cos2 a.
Mihaly Bencze
PP25122. Determine all x, y, z ∈ Z for which
sin 2x < 1y2(2−z2)
sin 2y < 1z2(2−x2)
sin 2z < 1x2(2−y2)
.
Mihaly Bencze
PP25123. If xk > 0 (k = 1, 2, ..., n) , thenn�
k=1
x2n−3k ≥ 1
2n−2 |(x1 − x2) (x2 − x3) ... (xn − x1)|+
+ n−1
���xn−11 x2...xn−1
� ��x1x
n−12 x3...xn−1
�...��
x1x2...xn−2xn−1n−1
�.
Mihaly Bencze
PP25124. In all triangle ABC holds� r3a
(tgA2+tgC
2 )2tg2 B
2
≥ (4R+r)3−12s2R36 .
Mihaly Bencze
PP25125. In all triangle ABC holds� ra
2rb+rc+ s2
3((4R+r)2−2s2)≥ 4
3 .
Mihaly Bencze
PP25126. In all triangle ABC holds� −a+b+c
3a−b+c + r(4R+r)3(s2−2r2−8Rr)
≥ 43 .
Mihaly Bencze
PP25127. If ak ∈ (0, 1] (k = 1, 2, ..., n) then determine all λ ∈ R for which
n�k=1
(1−ak)λ+1
aλk≤
�
1−n�
k=1ak
�λ+1
n�
k=1aλk
.
Mihaly Bencze
PP25128. If a, b > 0 then�a5 + 4a2b3 + b6
� �b5 + 4b2a3 + a6
�≥�
ab�a3 + 2b3
�+ b2
�2a3 + b3
�� �ab�b3 + 2a3
�+ a2
�2b3 + a3
��.
Mihaly Bencze
Proposed Problems 273
PP25129. If x ∈�0, π2�then
sinn x1+sinx+...+sinn x + cosn x
1+cosx+...+cosn x + (n−1)(sin x+cosx)(n+1) sinx cosx ≥ 2n
n+1 .
Mihaly Bencze
PP25130. Solve in R the following system
x21 + 2 {x2}2 = 5 [x3]2
x22 + 2 {x3}2 = 5 [x4]2
−−−−−−−−−x2n + 2 {x1}2 = 5 [x2]
2
where
[·] and {·} denote the integer, respective the fractional part.
Mihaly Bencze
PP25131. Compute limn→∞
1�0
1�0
ln (xn + ey) ln (yn + ex) dxdy.
Mihaly Bencze
PP25132. Determine all a, b, c ∈ N for which xm,n = (am)!(bn)!m!n!(m+n)! and
xm,n−1 + xm−1,n = cxm−1,n−1 for all m,n ∈ N∗.
Mihaly Bencze
PP25133. If n ∈ N , n ≥ 2 and 1 = d1 < d2 < ... < dk = n are the divisorsof n, then determine all r ∈ {1, 2, ..., n} for which n is prime if and only ifd1d2...dr + d2d3...dr+1 + ...+ dnd1...dr−1 = nk (r − 1) .
Mihaly Bencze
PP25134. In all scalene triangle ABC holds� m2a
(ma−mb)2(ma−mc)
2 ≥ 23(s2−r2−4Rr)
Mihaly Bencze
PP25135. Determine all a, b, c ∈ Z for which the equation�a2 + b2
�x2 − 2
�b2 + c2
�x−�c2 + a2
�= 0 have rational roots.
Mihaly Bencze
PP25136. In all triangle ABC holds��
tgA2 tg
B2 + 3
√3�
ctgB2
�ctgA
2 ctgC2 ≥ 28
√3.
Mihaly Bencze
274 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25137. Solve in Z the equation x1x2(1+x3)
+ x2x3(1+x4)
+ ...+ xnx1(1+x2)
= nk+1
when k ∈ Z\ {−1} is given.
Mihaly Bencze
PP25138. If λ = limn→∞
√n (2n−1)!!
(2n)!! then compute limn→∞
n�λ−√
n · (2n−1)!!(2n)!!
�.
Mihaly Bencze
PP25139. If xk ∈�0, π2�(k = 1, 2, ..., n) then
n�k=1
11−(sinxk)
n−1 ≥ �cyclic
11−sinx1 sinx2... sinxn−1
.
Mihaly Bencze
PP25140. Prove thatn�
k=1
����1
3√
k(k+1)− sin 1
3√
k(k+1)
���� <6nn+1 .
Mihaly Bencze
PP25141. In all triangle ABC holds�� b+c
2
�2cos2 A
2 ≤ 6�s2 − r2 − 4Rr
�.
Mihaly Bencze
PP25142. If xk > 0 (k = 1, 2, ..., n) then�
cyclic
�x1 −
√x1x2 + x2
�2 ≥n�
k=1
x2k.
Mihaly Bencze
PP25143. If xk > 0 (k = 1, 2, ..., n) then determine all p ∈ N for whichn�
k=1
xk ≥ p n
�1n
n�k=1
xnk + (n− p) n
�n�
k=1
xk.
Mihaly Bencze
PP25144. If λ > 0 and zk ∈ C∗ (k = 1, 2, ..., n) , thenn�
k=1
���zk + λzk
���4≥ 4nλ2 + 8λ
n�k=1
Re�z2k�.
Mihaly Bencze
Proposed Problems 275
PP25145. Determine all a, b, n,m ∈ N for which(an)! + (bm)! ≤ (n!)a+1 + (m!)b+1 .
Mihaly Bencze
PP25146. Determine all m,nk ∈ N (k = 1, 2, ..., n) for whichm�k=1
1((nk)!)
3 ≤ m�
n�
k=1nk
�
!.
Mihaly Bencze
PP25147. Let AkBkCk (k = 1, 2, ..., n) be triangles. Prove that
[·] and {·} denote the integer, respective the fractional part.
Mihaly Bencze
PP25151. Determine all polynomials P for which the zeros of P are inarithmetical progression, and the zeroes of P ′ are too in arithmeticalprogression.
Mihaly Bencze
276 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25152. Determine all polynomials P for which the zeros of P are ingeometrical progression, and the zeroes of P ′ are too in geometricalprogression.
Mihaly Bencze
PP25153. Determine all postve integers k, n for which the numberk (k − 1) ...10k (k − 1) ...10...k (k − 1) ...1 is prime.
Mihaly Bencze
PP25154. Let x be a nonzero real number sarisfying the equationa0x
n + a1xn−1 + ...+ an−1x+ an = 0, where a0, a1, ..., an ∈ Z such that
|a0|+ |a1|+ ...+ |an| > 1. Determine all n ∈ N∗ for which|x| > 1
|a0|+|a1|+...+|an| .
Mihaly Bencze
PP25155. Let ABC be a triangle. Determine all λ ∈ R for which�a2�
λra
− rrbrc
�= 4 (R+ (λ+ 1) r) .
Mihaly Bencze
PP25156. Consider n distinct points in the plane, n ≥ 3 arranged such thatthe number r (n) of segments of length d is maximed. Prove that
π2
2 − 154 <
∞�n=3
1r(n) .
Mihaly Bencze
PP25157. In all triangle ABC holds� 1
rstgB
2+2tgA
2tgC
2+2(tgA
2tgC
2 )2 ≥ s2
r(4R+r) .
Mihaly Bencze
PP25158. Solve in Z the following system:
�x+ y2
� �y2 + z
�= (z + x)3�
y + z2� �
z2 + x�= (x+ y)3�
z + x2� �
x2 + y�= (y + z)3
.
Mihaly Bencze
Proposed Problems 277
PP25159. Solve in C the following system
x4 − y + 1 = z2
y4 − z + 1 = x2
z4 − x+ 1 = y2.
Mihaly Bencze
PP25160. Determine all xk ∈ N∗ (k = 1, 2, ..., n) for which nx1x2...xn−1−11+x1x2...xn
isinteger.
Mihaly Bencze
PP25161. If a, b, c > 0 then� 1√
a3+2b3+6abc≤ 1√
abc.
Mihaly Bencze
PP25162. Denote pn the nth prime number. If n ≥ k (k + 1) then pn > knfor all k ∈ N∗, k ≥ 2.
Mihaly Bencze
PP25163. If a, b, c > 0 and� a2+1
a = 4 then
a√bc+ b
�c�1a + 1
b +1c
�+ c�a�1a + 1
b +1c
�+�1a + 1
b +1c
�√ab ≤
≤ 2�1 +�abc�1a + 1
b +1c
��.
Mihaly Bencze
PP25164. If xk ∈ R thenn�
k=1
|cosxk|+�
cyclic
|cos (x1 − x2)| ≥ n− 1.
Mihaly Bencze
PP25165. Compute limn→∞
n
�14 − ln
�12n
n�k=1
�2 + k
n2
���.
Mihaly Bencze
PP25166. In all triangle ABC holds� sin A
2
(cos A2 )
2 ≥ 2.
Mihaly Bencze
278 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25167. If b1, b2, ..., bn is a rearangement of numbers a1, a2, ..., an > 0 then
�1≤i<j≤n
aibjai+bj
≤ n
2n�
k=1ak
��
1≤i<j≤naibj
�.
Mihaly Bencze
PP25168. Let ABC be a triangle. Determine all λ > 0 for which� (a+b−c)λ
aλ+bλ−cλ≤ 3.
Mihaly Bencze
PP25169. Solve in C the following system
x21 +�x22 + 32 =
�x23 + 96
x22 +�x23 + 32 =
�x24 + 96
−−−−−−−−−−−−x2n +
�x21 + 32 =
�x22 + 96
Mihaly Bencze
PP25170. In all triangle ABC holds��a(b+c)
a2+bc≤�
2(5s2+r2+4Rr)s2+r2+2Rr
.
Mihaly Bencze
PP25171. Let n ≥ 2 be an integer. Find the number of integers k with0 ≤ k < n and such that k3 leaves a remainder of 1 when divided by n.
PP25173. If λ ∈ (0, 1) then determine all x ∈ R for which
�(a+ x) (b+ x) = x+
�aλ+bλ
2
� 1λ.
Mihaly Bencze
Proposed Problems 279
PP25174. If an+1 = [an] {an} for all n ≥ 0 when [·] and {·} denote theinteger part respectively the fractional part. Determine all a0 ∈ R for whichthe sequence (an)n≥0 is periodical.
Mihaly Bencze
PP25175. If M is a set for which cardM = 2n, then the number ofpartitions of M formed by 2-subsets of M is (2n)!
2n·n! .
Mihaly Bencze
PP25176. Let ABC be a triangle. Determine the best constant λ > 0 for
which�
R2r ≥ λ3a2b2c2
�
(λa2−(b−c)2).
Mihaly Bencze
PP25177. Prove that�2k+1
2k
�−�
2k
2k−1
�is divisible by 8k for all k ∈ N.
Mihaly Bencze
PP25178. Determine all functions f : (0,+∞) → (0,+∞) for which(f (x) + f (y) + 2xf (xy)) f (x+ y) = f (xy) for all x, y > 0.
Mihaly Bencze
PP25179. Solve in R the equationn�
k=1
�k3
x
�= n2(n+1)2
4 when [·] denote the
integer part.
Mihaly Bencze
PP25180. If x ∈�0, π2�then�
n�k=1
1+(sinx)4kn
1+(sinx)8kn
��n�
k=1
1+(cosx)4kn
1+(cosx)8kn
�< 1
sin2 x cos2 x.
Mihaly Bencze
PP25181. Solve in R the following equationn�
k=1
�kx
�= n(n+1)
2 when [·]denote the integer part.
Mihaly Bencze
280 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25182. Solve in R the following equationn�
k=1
�k2
x
�= n(n+1)(2n+1)
6 when
[·] denote the integer part.
Mihaly Bencze
PP25183. Compute λ = limn→∞
�n�
k=1
arctgkk2+1
− 12arctg
2n
�and
limn→∞
n
�λ+ 1
2arctg2n−
n�k=1
arctgkk2+1
�.
Mihaly Bencze and Lajos Longaver
PP25184. In all triangle ABC holds� √
sinA+√sinB
�
cos A−B2
cos C2+�
|sin A−B2 | sin C
2
≤ 6.
Mihaly Bencze
PP25185. In all triangle ABC holds−195
8 ≤ 3�
cos 2A− 8�
sinA cosB + 5(2R−s+2r)R ≤ 60.
Mihaly Bencze and Lajos Longaver
PP25186. Ifn�
k=1
|shkx| ≥ n(n+1)2 |shx| .
Mihaly Bencze
PP25187. If xk > 0 (k = 1, 2, ..., n) andn�
k=1
xk = 1 thenn�
k=1
xarctgxkk ≥ 1.
Mihaly Bencze
PP25188. In all acute triangle ABC holds��√tgA+ 4ctgB +
√tgB + 4ctgA+
√tgA+ 9ctgB +
√tgB + 9ctgA
�≥
≥ 21√2.
Mihaly Bencze
PP25189. In all acute triangle ABC holds
2�
tg2A2 tg2B2 ≥ 1 +
�tg2B2
�12
�tg4A2 + tg4C2
�.
Mihaly Bencze
Proposed Problems 281
PP25190. Solve the following system
x41 + 35x22 + 24 = 10x3�x24 + 5
�
x42 + 35x23 + 24 = 10x4�x25 + 5
�
−−−−−−−−−−−−−−x4n + 35x21 + 24 = 10x2
�x23 + 5
�.
Mihaly Bencze
PP25191. If x ∈ R then 2ex
e2x+1+ e−x
�e4x+1
2 ≤ (ex+1)2
2ex +((x−1)e2x+x+1)
2
2x2ex(e2x+1).
Mihaly Bencze
PP25192. Prove thatn�
k=1
k
(k+2)�
k√1·2+ k√2·3+...+ k√
k(k+1)� ≥ n
n+3 .
Mihaly Bencze
PP25193. In all acute triangle ABC holds
1).� 1
1−sinA sinB ≤�
s2+r2−4R2
s2−(2R+r)2
�2− 8R(R+r)
s2−(2R+r)2
2).� 1
1−cosA cosB ≤�s2+r2+Rr
2sr
�2− 4R
r
Mihaly Bencze
PP25194. If ak > 0 (k = 1, 2, ..., n) andn�
k=1
ak = 1 then
n�k=1
�1 + 1
ak
�≥ (n+ 1)n .
Mihaly Bencze
PP25195. Determine all a, b, c > 0 for which (a+ b+ c)�1a + 1
b +1c
�≤ 81
8 .
Mihaly Bencze
PP25196. Let ABCD be a convex quadrilateral. Determine min and maxof the following expression
�sin A
2 cos B2 tg
C4 .
Mihaly Bencze
PP25197. If ak, bk, ck ∈ N\ {0, 1} thenn�
k=1
�1− 1
a2k
�k �1− 1
b2k
�k �1− 1
c2k
�k> 8n−1
7·8n .
Mihaly Bencze
282 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25198. In all acute triangle ABC holds
R+rR <
� 1A2
�2A�A
sinxx dx
��2A�A
xdxsinx
�< s2+r2−4R2
2(s2−(2R+r)2).
Mihaly Bencze
PP25199. Solve in N the equation xy + yx = (x− y)2(x+y) .
Mihaly Bencze
PP25200. Solve in R the following system
�1 + 3
1x
�(13− 7y) = 16 (2 + 7z)�
1 + 31y
�(13− 7z) = 16 (2 + 7x)�
1 + 31x
�(13− 7x) = 16 (2 + 7y)
.
Mihaly Bencze and Gyorgy Szollosy
PP25201. Solve in N the equationn�
k=1
(1 + xk) = ynn�
k=1
xk.
Mihaly Bencze
PP25202. Solve in R the following system
2x2 + 3 ≥ 3y + 2√z2 − z + 1
2y2 + 3 ≥ 3z + 2√x2 − x+ 1
2z2 + 3 ≥ 3x+ 2�y2 − y + 1
.
Mihaly Bencze
PP25203. If zp ∈ C∗ (p = 1, 2, ..., n) such that |z1| = |z2| = ... = |zn| anda1 =
z1+z2z1−z2
, a2 =z2+z3z2−z3
, ..., an = zn+z1zn−z1
then�a1a2...a2k +
�a1a2...a2k−2 + ...+ 1 = 0 if n = 2k + 1 and�
a1a2...a2k−1 +�
a1a2...a2k−3 + ...+�
a1 = 0 if n = 2k.
Mihaly Bencze
PP25204. If a > 1 and xk > 0 (k = 1, 2, ..., n) then
PP25217. If a, b, c > 0 then determine all λ ∈ R for which� a
a+λb+c ≥ λ+1λ+2 .
Mihaly Bencze
PP25218. If x, y, z > 0 and x+ y + z = 1 then� x1−x4 + 81
80 ≥ 16� 1−x
16−(1−x)4.
Mihaly Bencze
PP25219. If x, y, z > 0 and x+ y + z = 1, then� x1−x4 + 81
80 ≥ 27� 2x+y
81−(2x+y)4+ 27
� x+2y
81−(x+2y)4.
Mihaly Bencze
PP25220. If x, y, z > 0 and x+ y + z = 1, then� x1−x4 + 27
� 2x+y
81−(2x+y)4≥ 54
� x+2y
81−(x+2y)4.
Mihaly Bencze
Proposed Problems 285
PP25221. If x, y, z > 0 and x+ y + z = 1, then� x1−x4 + 64
� 1+x256−(1+x)4
≥ 16� x+y
16−(x+y)4.
Mihaly Bencze
PP25222. Solve in R the following system:� √x4 + 1 +
�y4 + 1 +
√z4 + 1 = 3
√2
xy + yz + zx = 1
Mihaly Bencze
PP25223. If a, b, c ∈ C then
3�|a|2 + |b|2 + |c|2 + |a+ b+ c|2
�≥ (|a+ b|+ |b+ c|+ |c+ a|)2 .
Mihaly Bencze
PP25224. If a, b, c ∈ C then
4�|a+ b|2 + |b+ c|2 + |c+ a|2
�≥ (|a|+ |b|+ |c|+ |a+ b+ c|)2 .
Mihaly Bencze
PP25225. If ak ∈ [0, 1] (k = 1, 2, ..., n) then�
cyclic
a11+a2+a3+...+an
+n�
k=1
(1− ak) ≤ 1.
Mihaly Bencze
PP25226. If x, y, z > 0 and λ ≥ 1 then��
1 + x1λ
�λ+ 3
�1 +�x+y+z
3
� 1λ
�λ
≥ 2��
1 +�x+y
2
� 1λ
�λ
.
Mihaly Bencze
PP25227. If ak > 1 (k = 1, 2, ..., n) (n ≥ 3) then solve in R the following
system:
n�k=1
axk =
�n�
k=1
ak − n
�y + n
n�k=1
ayk =
�n�
k=1
ak − n
�z + n
n�k=1
azk =
�n�
k=1
ak − n
�x+ n
.
Mihaly Bencze
286 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25228. If a, b, c ∈ C are given, then solve the following system:(x1 + a)x2 + b = (x2 + a)x3 + b = ... = (xn + a)x1 + b = c.
Mihaly Bencze
PP25229. Let ABC be a triangle and MNK be the triangle for whichMN = ma, NK = mb, KM = mc. Let T ∈ Int (MNK) . Prove that(TM + TN + TK) (TM · TN + TN · TK + TK · TM) ≥≥ TM ·m2
a + TN ·m2b + TK ·m2
c .
Mihaly Bencze
PP25230. Let A1A2...An be a convex polygon, and M ∈ Int (A1A2...An) .
Prove that 12
� MA21+MA2
2−A1A22
MA1·MA2+ n
n−1 ≥ 0.
Mihaly Bencze
PP25231. Denote x1 < 0 < x2 the roots of equationx4 + a1x
3 + b1x2 + c1x+ d1 = 0. Determine all ak, bk, ck, dk ∈ C (k = 1, 2) for
which x1x2 is a root of the equation x6 + x4 + a2x3 + b2x
2 + c2x+ d2 = 0.
Mihaly Bencze
PP25232. Let ABC be a triangle. Solve in R the following system:
�x
(x2+1) cos A2
= y
(y2+1) cos B2
= z(z2+1) cos C
2
xy + yz + zx = 1
Mihaly Bencze
PP25233. Let ABC be a triangle with sides a, b, c. Solve in R the followingsystem:
�x
(x2+1)a= y
(y2+1)b= z
(z2+1)c
xy + yz + zx = 1.
Mihaly Bencze
PP25234. In all triangle ABC holds 8 (�
ma) (�
mamb) ≥ 27�
a2ma.
Mihaly Bencze
Proposed Problems 287
PP25235. If x0, y0, z0 > 0 and 2xn+1 = xn + 1yn; 2yn+1 = yn + 1
zn,
2zn+1 = zn + 1xn
for all n ≥ 1 then the sequences (xn)n≥1 , (yn)n≥1 and(zn)n≥1 are convergent and compute its limits.
Mihaly Bencze
PP25236. If ak, bk ∈ R (k = 1, 2, ..., n) and c ≥n�
k=1
(ak − bk)2 then
vs�n�
k=1
a2k
��n�
k=1
b2k − c
�+
�n�
k=1
b2k
��n�
k=1
a2k − c
�≤ 2
�n�
k=1
akbk
�2
.
Mihaly Bencze
PP25237. If ak ≥ −1 (k = 1, 2, ..., n) , then 9n�
k=1
a3k +3n�
k=1
a2k + n ≥ 5n�
k=1
ak.
Mihaly Bencze
PP25238. Determine all triangles ABC and all points M ∈ (BC) andK ∈ (BC) such that
�AB2 ·MC +AC2 ·MB
� �AB2 ·DK +AC2 ·DB2
�=
=�AB2 ·MC2 +AC2 ·MB2
� �AB2 ·KC +AC2 ·KB
�.
Mihaly Bencze
PP25239. If ak > 0 (k = 1, 2, ..., n) andn�
k=1
ak = 1 then
nn�
k=1ak
≤ �cyclic
11+a1+a2+...+an−1
≤ 1.
Mihaly Bencze
PP25240. If ak > 0 (k = 1, 2, ..., n) andn�
k=1
ak = n then
n�k=1
n−1−a3kak
≥ n (n− 2) .
Mihaly Bencze
288 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25241. If ak > 0 (k = 1, 2, ..., n) then
1).n�
k=1
ak +�
cyclic
a1(a2+...+an)
2 ≥ 2nn−1
2).n�
k=1
a2k +�
cyclic
a21(a2+...+an)
4 ≥ 2n(n−1)2
Mihaly Bencze
PP25242. Prove that12
n�k=1
1k2
+ 12
n�k=1
�12 + 3
4 + ...+ kk+1
�2+
n�k=1
1k
�12 + 1
3 + ...+ 1k+1
�≥ n.
Mihaly Bencze
PP25243. If n, k ∈ N∗ then
12k
+ 13k
+ ...+ 1(n+1)k
≤ n
�1− n
�(2k−1)(3k−1)...((n+1)k−1)
((n+1)!)k
�.
Mihaly Bencze
PP25244. In all triangle ABC holds�
bc�1 + sin A
2
�2 ≤ s2 + 6r2 + 24Rr.
Mihaly Bencze
PP25245. In all triangle ABC holds�
bc sin A2 ≤ 2r (4R+ r) .
Mihaly Bencze
PP25246. In all triangle ABC holds�
(a− b)2 + 3�abc�1 + sin A
2
� �1 + sin B
2
� �1 + sin C
2
�� 23 ≤ 3s2.
Mihaly Bencze
PP25247. If ak ∈ N∗ (k = 1, 2, ..., n) such thatn�
k=1
a2k is a perfect square,
then exist p, q, r ∈ N∗ for which
n�
k=1ak
�
1≤i<j≤n
aiaj
2
= 1p + 2
q +1r .
Mihaly Bencze
Proposed Problems 289
PP25248. If ak ∈ N∗ (k = 1, 2, ..., n) such thatn�
k=1
a2k is a perfect square,
then exist br ∈ N∗ (r = 1, 2, ...,m+ 1) such that
n�
k=1ak
�
1≤i<j≤n
aiaj
m
=m+1�r=1
1br.
Mihaly Bencze
PP25249. If x0 = x1 = 1 and n (n+ 1)�xn+1
√xn−1 − xn
√xn�=
√xn−1xn
for all n ≥ 1, then compute limn→∞
n (4− xn) .
Mihaly Bencze
PP25250. In all triangle ABC holds
1).��
1 + sin A2
�2 ≤ s2
2Rr
2).��
1 + sin A2
�4 ≤ s2(s2−r2−4Rr)8R2r2
3).�
sin A2 ≤ s2+r2+4Rr
4Rr
4).��
1 + sin A2
�≤ s2
4Rr
Mihaly Bencze
PP25251. If a, b, c > 0 then 3�
a2�
ab ≥� 1a+b +
16
��� c(a−b)2
(a+c)(b+c)�
ab
�2
.
Mihaly Bencze
PP25252. In all triangle ABC holds� b2+c2
(2a2+2c2−b2)(2a2+2b2−c2)≤ s2−r2−4Rr
9s2r2.
Mihaly Bencze
PP25253. Determine all ak ∈ N∗ (k = 1, 2, ..., n) for whichn�
k=1
3k2+3k+1a3ka
3k+1
=n(n2+3n+3)
(n+1)3.
Mihaly Bencze
PP25254. Solve in R the following system
2[log2 x1] + x[log2 x3]2 = 2[log2 x2] + x
[log2 x4]3 = ... = 2[log2 xn] + x
[log1 x2]2 = 2 where
[·] denote the integer part.
Mihaly Bencze
290 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25255. Prove thatn−1�k=1
1cosx−cos kπ
n
= cosxsin2 x
− n cos(nx)sinx sinnx .
Mihaly Bencze
PP25256. Let a0, a1, ..., an−1 ∈ N∗ such that a0 + a1 + ...+ an−1 = n. Provethat k0a0 + k1a1 + ...+ kn−1an−1 =
kn−1k−1 if and only if
a0 = a1 = ... = an−1 = 1, when k ≥ 2.
Mihaly Bencze
PP25257. Let 1 = d1 < d2 < ... < dk = n the positive divisors of n. Provethat d1, d2, ..., dk are not in geometrical progression.
Mihaly Bencze
PP25258. In all triangle ABC holds r�s2 − (2R+ r)2
�≤ 4R3.
Mihaly Bencze
PP25259. Determine all functions f : N → N for which
f�x4 − x3f (y) + x2f2 (y)− xf3 (y) + f4 (y)
�=
x5+f(y5)x+y for all x, y ∈ N.
Mihaly Bencze
PP25260. Prove thatn�
k=1
1
(k+2)((2k)!)1k≥ n(n+3)
2(n+1)(n+2) .
Mihaly Bencze
PP25261. Determine all bijective functions f : [0,+∞) → [0,+∞) forwhich f (x+ y + z) =f (x)+ f (y)+ f (z)+2f−1 (f (x) f (y))+2f−1 (f (y) f (z))+2f−1 (f (z) f (x))for all x, y, z ≥ 0.
Mihaly Bencze
PP25262. Solve in Z the equation x3 + y3 = 7z + 2016.
PP25264. Determine all xk ∈ N (k = 1, 2, ..., n) for whichn�
k=1
x2k +n�
k=1
xk =n�
k=1
xk.
Mihaly Bencze
PP25265. If xn+m + xn−m = x4n for all n,m ∈ N (n ≥ m) then determinex2016.
Mihaly Bencze
PP25266. In all triangle ABC holds
1).� a+b
b+c ≥ 134 − 3r(r+4R)
4s2
2).� a
b ≥ 4− 3r(4R+s)s2
3).� ra+rb
rb+rc≥ 4− 3s2
(4R+r)2
4).� sin2 A
2+sin2 B
2
sin2 B2+sin2 C
2
≥ 4− 3(s2+r2−8Rr)4(2R−r)2
5).� cos2 A
2+cos2 B
2
cos2 B2+cos2 C
2
≥ 4− 3(s2+(4R+r)2)4(4R+r)2
Mihaly Bencze
PP25267. Prove thatn�
k=1
�2k (k + 1)
√6�> n
√6
3(n+1) where {·} denote the
fractional part.
Mihaly Bencze
PP25268. Solve the equationn�
k=1
�x
k(k+1)(k+2)
�= n, where [·] denote the
integer part.
Mihaly Bencze
292 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25269. In all triangle ABC holds������
a cosA b cosB c cosCa b c1 1 1
������= s
2Rr (a− b) (b− c) (c− a) .
Mihaly Bencze
PP25270. In all triangle ABC holds:1).�
a ma = 2�s2 − r2 − 4Rr
�
2).�
ma wa = s2
3).�
ha wa = 4s2r2��
s2+r2+4Rr4sRr
�2− 1
Rr
�
Mihaly Bencze
PP25271. Solve in R the following system
x21 + 8 = 6x2 +�1 +
√x3 − 2
x22 + 8 = 6x3 +�1 +
√x4 − 2
−−−−−−−−−−−−−x2n + 8 = 6x1 +
�1 +
√x2 − 2
Mihaly Bencze
PP25272. Determine all z ∈ C∗ for which��z + 1
z
�� =��z2n+1 + 1
z2n+1
�� = r ≥ 2
Mihaly Bencze
PP25273. Determine all ak > 0 (k = 1, 2, ..., n) for which holdsn�
k=1
akxakx ≥ n for all x > 0.
Mihaly Bencze
PP25274. Determine all n ∈ N (n ≥ 2) for whichn�
k=1
n
�8 + 1
k > n�
n√5 + n
√7− n
√4�.
Mihaly Bencze
PP25275. Determine all 0 < x1 ≤ x2 ≤ ... ≤ xn for whicharctgx1, arctgx2, ..., arctgxn are in arithmetical progression.
Mihaly Bencze
Proposed Problems 293
PP25276. In all triangle ABC holds� 1
a3
�128R4s2r2 − a4 (b2 + c2 − a2)2 ≥ 3
√2srR .
Mihaly Bencze
PP25277. In all triangle ABC holds� 1
1+sin2 A≤� 1
1+sinA sinB .
Mihaly Bencze
PP25278. In all acute triangle ABC holds� 1
1+cos2 A≤� 1
1+cosA cosB .
Mihaly Bencze
PP25279. Let be 1 = d1 < d2 < ... < dk = n the positive divisors of n.Determine all n for which k = 2016 and n = d21 + d22 + ...+ d22016.
Mihaly Bencze
PP25280. If ak ∈ R (k = 1, 2, ..., n) such that aiaj ≥ 1 (i, j ∈ {1, 2, ..., n})then
n�k=1
1a2k+1
≥ �cyclic
1a1a2+1 . If 0 < aiaj ≤ 1 (i, j ∈ {1, 2, ..., n}) then holds
the reverse inequality.
Mihaly Bencze
PP25281. Let ABCD be a convex quadrilateral, we construct in exteriorthe equilateral triangles ABE,BCF,CDG and DAH. Determine allquadrilaterals ABCD for which the lines AC,BD,EG,FH are concurents.
Mihaly Bencze and Ferenc Olosz
PP25282. If a, b > 1 then solve in R the following systemax1 + b−x2 = ax2 + b−x3 = ... = axn + b−x1 = a+ b
Mihaly Bencze and Gyorgy Szollosy
PP25283. In all triangle ABC holds�� a
ma
�λ≥ 3�
2√3
�λfor all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
294 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25284. If a > 1 and b ≥ −14 then solve in (0,+∞) the following system
PP25354. In all triangle ABC holds� (a3+2b3+c3+abc)(a3+c3+abc)
(a3+b3+abc)(b3+c3+abc)≤ 3(s2−3r2−6Rr)
2Rr .
Mihaly Bencze
PP25355. 1). If a > 0, a �= 1, b > 1 and n ∈ N∗ then compute
In (a, b) =b�1b
((1−n)x+ln a)xn−3a1x dx
xn−1a1x+1
2). Determine all continuous functions g :�12 , 2�→ R for which
2�12
�(2−x)ex
x(x2+ex)g (x) dx ≥ ln
4(4√e+1)
e2+4≥
2�12
g2 (x) dx.
Traian Ianculescu
304 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25356. 1). If f : [a, b] → R is continuous and a > 0, α ∈ R, α �= 1, then
exist c ∈ (a, b) such that (α− 1) (ab)α−1b�af (x) dx =
�bα−1 − aα−1
�cαf (c)
2). If g : [a, b] → R is a continuous function and G (x) =x�0
g (t) dt, x ∈ [a, b]
and β < 0 then exists c1, c2 ∈ (0, b) , c1 �= c2 such that
1−βb1−β
b�0
G (x) dx = cβ+11 g (c2)
Traian Ianculescu
PP25357. If 0 < a < 1 < c < d then determine all differentiable functionwith continuous derivate for whichf (a · cx + b)− f (a · dx + b) = a (f (cx)− f (dx)) for all x ∈ R.
Mihaly Bencze
PP25358. Solve in Z the equation 2x
y+1 + 3y
2x+1 = 2.
Mihaly Bencze
PP25359. If a0 =13√4
and an+1
�1 + a2n
�= an for all n ≥ 1 the
a242 ∈�
110 ,
19
�and lim
n→∞3√nan = 1
3√3
Traian Ianculescu
PP25360. Prove thatn�
k=1
�2n+2−k
k ≥ n�1 +
n√n!
n+1
�.
Traian Ianculescu
PP25361. If A ∈ M2 (R) and x ∈ R such that det�A2 − x2I2
�≥ 0 then
1). If detA+ x2 ≥ 0 thendet (An − xnI2) + det (An + xnI2) ≥ 22−n |xTr (A)|n for all n ∈ N
PP25405. If ap > 1 (p = 1, 2, ..., n) and k ∈ {1, 2, ..., n} thenlog2a1+a2+...+ak
(a2+a3+...+ak+1)
a3+a4+...+ak+2+
log2a2+a3+...+ak+1(a3+a4+...+ak+2)
a4+a5+...+ak+3+ ...
+log2an+a1+...+ak−1
(a1+a2+...+ak)
a2+a3+...+ak+1≥ n2
kn�
p=1ap
.
Mihaly Bencze
Proposed Problems 311
PP25406. If (xn)n≥1 is a real sequence such that |xn+1 − xn| ≤ 1 andnyn = x1 + x2 + ...+ xn for all n ∈ N∗, then |yn+1 − yn|+ |yn+2 − yn+1| ≤ 1for all n ∈ N∗.
Mihaly Bencze
PP25407. If An =
������
cos2 nx cos2 ny cos2 nzsin 2nx sin 2ny sin 2nzcos 2nx cos 2ny cos 2nz
������where x, y, z ∈ R then
n�k=1
|Ak| < 2n.
Mihaly Bencze
PP25408. Solve in N the equation�2n
n
�= 2k, when [·] denote the integer
part.
Mihaly Bencze
PP25409. Solve in N the equation�3n
n
�= 3k, when [·] denote the integer
part.
Mihaly Bencze
PP25410. Solve in N the equation�2n+3m
nm
�= 6k, when [·] denote the
integer part.
Mihaly Bencze
PP25411. Let a1, a2, ..., an2 be an arithmetical progression, and let be the
n elements from this tabel in the rest n (n− 1) elements exist n numbers inarithmetical progression.
Mihaly Bencze
312 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25412. If xk ∈ R (k = 1, 2, ..., n) such thatn�
k=1
sinxk = 0, then
n�k=1
|cosx− sinxk| ≤ n for all x ∈ R.
Mihaly Bencze
PP25413. Denote An the set of every integers of form ±1± 2± ...± n. Byexample A2 = {−3,−1, 1, 3} and A3 = {−6,−4,−2, 0, 2, 4, 6} . Denote |An|the number of elements of the set An. Compute
∞�n=1
1|An| .
Mihaly Bencze
PP25414. If a, b ∈ C then (|1 + ab|+ |a+ b|)2 ≥���(1 + ab)2 − (a+ b)2
��� .
Mihaly Bencze
PP25415. If A ∈ M2 (R) then83 det
�A2 +A+ I2
�≥ (1− detA)2 + (1 + TrA)2 .
Mihaly Bencze
PP25416. If A ∈ Mn (R) is an antisimmetric matrice(aij + aji = 0 for all i, j ∈ {1, 2, ..., n}) then for all xk ≥ 0 (k = 1, 2, ...,m)
holdsm�k=1
det (A+ xkIn) ≥�det
�A+ m
�m�k=1
xkIn
��m
.
Mihaly Bencze
PP25417. In all triangle ABC holds�� ab
(b+c)(c+a) ≤�
6(s2−r2−Rr)s2+r2+2Rr
.
Mihaly Bencze
PP25418. In all triangle ABC holds s2+(4R+r)2
4sR − 3s2(4R+r) ≥
√3.
Mihaly Bencze
PP25419. Solve in Z the equation 1x + 1
y+2 + 1z+3 = 3.
Mihaly Bencze
Proposed Problems 313
PP25420. If x1 = 2 and x2n+1 = xn + 1n for all n ≥ 1 then compute
limn→∞
n (1− xn) .
Mihaly Bencze
PP25421. In all triangle ABC holds� a(a+b)3(a+c)
(b+c)(b(a+b)3+(a+c)c3+7(a+c)(a+b)3)≤ 1
3 .
Mihaly Bencze
PP25422. Prove that
n�
k=1(−1)k−1 1
k2(nk)
Hn
Hn
≥n�
k=1
(Hk)1k when Hn =
n�k=1
1k .
Mihaly Bencze
PP25423. Let G = (0,+∞) and (log3 (x ◦ y))n = (log3 x)n + (log3 y)
n − 3n
where n = 2k + 1, k ∈ N∗ and for all x, y ∈ G
1). Prove that (G, ◦) is Abelian group
2). Prove that (G, ◦) ∼= (R,+)
Mihaly Bencze
PP25424. If Hn = 1 + 12 + ...+ 1
n then
n�k=1
�1 + 1
k
�nk
k+1�0
xn−1 ln�
kk+1 − x
�dx = − (Hn + ln (n+ 1)) .
Mihaly Bencze
PP25425. If ak > 0 (k = 1, 2, ..., n), then
�cyclic
a1
�1aλ2
+ 1aλ3
+ ...+ 1aλn
�≥
n�
k=1ak
(n−1)λ−1
n�
k=1
1ak
− nn�
k=1
ak
λ
for all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP25426. If a > 1, b ≥ −14 then solve in (0,+∞) the equation
a−bxloga x + log2a x = x+ logb x+ b.
Gyorgy Szollosy
314 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25427. If Hn = 1 + 12 + ...+ 1
n thenn�
k=0
Hkk ≥ 1
2 (Hn)2 + ln2 n
2n .
Mihaly Bencze
PP25428. Solve in R the equation 9x2 − 9x− 2 = 12 sinπx.
Gyorgy Szollosy
PP25429. Solve in R the equation�65
� 1x + 2
1+( 95)
x−1 = 3.
Gyorgy Szollosy
PP25430. If n,m ∈ N (n ≥ 2) , α, ak > 0 (k = 1, 2, ..., n) such thatn�
k=1
aαk ≤ n then 1n + m
n�
k=1ak
≥ 1+mnn�
k=1ak
.
Gyorgy Szollosy
PP25431. Solve in R the equation 3x + 35y = 2y − 2
3x = 5.
Gyorgy Szollosy
PP25432. In all scalene triangle ABC holds�� b−a
ma−mb− 2
3
�2+
32(s2+r2−2Rr)s2+r2+2Rr
< 64�
s2−r2−Rrs2+r2+2Rr
�2.
Mihaly Bencze
PP25433. In all scalene triangle ABC holds�� b−a
ma−mb− 2
3
��c−b
mb−mc− 2
3
�≤ 16(s2+r2−2Rr)
s2+r2+2Rr.
Mihaly Bencze
PP25434. In all scalene triangle ABC holds�b−a
ma−mb− 2
3
��c−b
mb−mc− 2
3
��a−c
mc−ma− 2
3
�≤ 128Rr
s2+r2+2Rr.
Mihaly Bencze
PP25435. In all scalene triangle ABC holds�� b−a
ma−mb
�2≥ 16
27
�� ma+mba+b
�2.
Mihaly Bencze
Proposed Problems 315
PP25436. In all scalene triangle ABC holds� b−a
ma−mb<
2(7s2−r2+2Rr)3(s2+r2+2Rr)
.
Mihaly Bencze
PP25437. Denote F the area of the triangle with sides cos A2 , cos
B2 , cos
C2
when ABC is a given triangle. Prove that
16F 2 ≤ cos A2 cos B
2 cos C2
�cos A
2 + cos B2 + cos C
2
�.
Mihaly Bencze
PP25438. In all triangle ABC holds3�s2 − r2 − 4Rr
�2 − 12s2r2 ≥�
mamb (a− b)2 .
Mihaly Bencze
PP25439. In all triangle ABC holds� (a+b)2m2
b+(a+c)m2c
2a+b+c ≥�mamb.
Mihaly Bencze
PP25440. In all triangle ABC holds7(s2−r2−4Rr)
4 ≥� ama ≥ s(s2+r2−2Rr)2R .
Mihaly Bencze
PP25441. In all triangle ABC holds 3�
(2a+ b+ c) (mbmc)2 +
916
�(b+ c) (b+ c− a) (b− c)2
�a2 + b2 + c2 + 3a (b+ c)
�≥
≥��
(a+ b)m2b +�
(a+ c)m2c
�2.
Mihaly Bencze
PP25442. In all triangle ABC holds
6�
(b− c) (bmc − cmb) (bmc + cmb) ≥���
(b+ c) (a2 + b2 + c2) |b− c|�2
.
Mihaly Bencze
PP25443. If Sn =n�
k=1
2k−1�
k2
k+1
�when [·] denote the integer part, then
�2−n−1Sn
�= n
2 − 1 for all n = 2k.
Mihaly Bencze
316 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25444. Let A1A2...An be a convex polygon. Prove that
n+�
cyclic
�a1
a2+...+an
�1 +�
a2+...+ana2+...+an−a1
���=�
cyclic
�a2+...+an
a2+...+an−a1
�when [·]
denote the integer part.
Mihaly Bencze
PP25445. Let ABCD be a convex quadrilateral such that A∡ ≥ 90◦,C∡ ≥ 90◦. Denote E respective F the proiection of A and C to BD. Provethat 1
AE·CF ≥ 1AB·DC + 1
AD·BC .
Mihaly Bencze
PP25446. Prove thatm�
n=1
1n�
k=1[√k2−k+1+
√k2+k+1]
= mm+1 .
Mihaly Bencze
PP25447. Let a, b, c > 0 then determine all λ ∈ R for which� ab ≥ λ
� abbλ+ca
.
Mihaly Bencze
PP25448. In all triangle ABC holds� a+b
c
��b+c−a
a +�
c+a−bb
�≤�s2+r2−2Rr
2Rr
�2− 2(s2−r2−Rr)
Rr .
Mihaly Bencze
PP25449. If a1 = 1 and ak > 0 (k = 1, 2, ..., n) such that
1a1+a2
+ 1a2+a3
+ ...+ 1an−1+an
= an − 1 for all n ≥ 1 thenn�
k=1
1a2ka
2k+1
= nn+1 .
Mihaly Bencze
PP25450. In all triangle ABC holds
1).��
a2 + b2
3 + c2�≥ 8s2Rr
�s2 + r2 + 2Rr
�
2).��
r2a +r2b3 + r2c
�≥ 4s4Rr2
3).���
sin A2
�4+ 1
3
�sin B
2
�4+�sin C
2
�4� ≥ r2((2R−r)(s2+r2−8Rr)−2Rr2)512R5
Mihaly Bencze
Proposed Problems 317
PP25451. Let A1A2...An be a convex polygon. Prove that�
cyclic
√an(a1+a2+...+an−1−an)
a1+a2+...+an−1≤ n
2 .
Mihaly Bencze
PP25452. In all triangle ABC holds� r4a
λr2br2c+((4R+r)2−2s2)
2 ≥ 3λ+3 for all
λ > 0.
Mihaly Bencze
PP25453. If ak > 0 (k = 1, 2, ..., n) and λ > 0 then
�cyclic
1
λa1+2n�
k=1ak
≤ 1(n+1)2
�1λ
n�k=1
+n�
cyclic
1a1+a2
�.
Mihaly Bencze
PP25454. In all triangle ABC holds� rarb
5ra+2(ra+rc)≤ 4R+r
12 + s2+r(4R+r)64R .
Mihaly Bencze
PP25455. In all triangle ABC holds� (s−a)(s−b)
5(a+c)−b ≤ s24 +
r(s2+(4R+r)2)8sR .
Mihaly Bencze
PP25456. If ak, bk ∈ R (k = 1, 2, ..., n) , then�
n�k=1
a2k
��n�
k=1
b2k
�≥�
n�k=1
akbk
�2
+ 2n(n−1)
��
1≤i<j≤n|aibj − ajbi|
�2
.
Mihaly Bencze
PP25457. If ak ∈ [0, 1] (k = 1, 2, ..., n) then�
cyclic
a1a31+2(a2+...+an)+n+1
≤ 13 .
Mihaly Bencze
PP25458. If x, ak > 1 (k = 1, 2, ..., n) , then loga1a2...an x ≤ n2n�
k=1
logak x.
Mihaly Bencze
318 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25459. Determine an (n ≥ 1) if a1 = 1 and1
a1a3a5+ 1
a3a5a7+ ...+ 1
a2n−1a2n+1a2n+3= n(n+2)
3a2n+1a2n+3.
Mihaly Bencze
PP25460. Determine an (n ≥ 1) if a1 = 1 and
14
a1a3+ 24
a3a5+ ...+ n4
a2n−1a2n+1=
n(n+1)(n2+n+1)6a2n+1
.
Mihaly Bencze
PP25461. Determine all function f : N → N for which(f(n3)+1)(f3(n)−1)
(f(n2)−n+1)(f2(n)+n+1)= f (n− 1) (f (n) + 1) for all n ∈ N∗.
Mihaly Bencze
PP25462. In all triangle ABC holds1).� a2
a4+b2c2≤ 1
4R
2).� r2a
r4a+r2br2c≤ 4R+r
2s2r
3).� (sin A
2 )4
(sin A2 )
8+(sin B
2sin C
2 )4 ≤ 4R(2R−r)
r2
4).� (cos A
2 )4
(cos A2 )
8+(cos B
2cos C
2 )4 ≤ 4R(4R+r)
s2
Mihaly Bencze
PP25463. If ak > 0 (k = 1, 2, ..., n) , then
�cyclic
(n−1)an−11 +a2a3...an
an−22
≥ n2 n
�n�
k=1
ak.
Mihaly Bencze
PP25464. Solve in R the following system:
[x1] +�x2 +
13
�+�x3 +
23
�= [3x4]
[x2] +�x3 +
13
�+�x4 +
23
�= [3x5]
−−−−−−−−−−−−−−−[xn] +
�x1 +
13
�+�x2 +
23
�= [3x3]
, when [·] denote the integer part.
Mihaly Bencze
Proposed Problems 319
PP25465. Solve in R the equation�x2�+�x+ 1
2
�= [2
√x] , when [·] denote
the integer part.
Mihaly Bencze
PP25466. If a1 = 0 and (an+1 − an) (an+1 − an − 2) = 4an for all n ≥ 1
thenn�
k=2
1ak
= n−1n .
Mihaly Bencze
PP25467. Solve in R the following system
�x21 − x2 − 2
�= [x3]�
x22 − x3 − 2�= [x4]
−−−−−−−−−�x2n − x1 − 2
�= [x2]
, when
[·] denote the integer part.
Mihaly Bencze
PP25468. Solve in R the following equation
n {x}+n�
k=1
�x+ 1
k
�= 1 +
n�k=1
{kx} , when {·} denote the fractional part.
Mihaly Bencze
PP25469. If Sn =n�
k=1
�k3
�when [·] denote the integer part, then compute
∞�n=3
1S2n.
Mihaly Bencze
PP25470. Determine all a, b, c, d ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for which��ab�2n −
�cd�2n���
cd�2n
+�cd�2n+1
+ 1�is divisible by abcd.
Mihaly Bencze
PP25471. If a1 =16 and (n+ 3) an+1 = (n+ 1)
�an + 1
2
�for all n ≥ 1, then
computen�
k=1
[ak] when [·] denote the integer part.
Mihaly Bencze
320 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25472. In all triangle ABC holds� 4
�
tgA2tgB
2
1+ 4�
9tgB2tgC
2
≥ 16 .
Mihaly Bencze
PP25473. Determine all ak ∈ R (k = 1, 2, ..., n) for which a1 = 1 andn�
k=1
a3k =
�n�
k=1
ak
��n�
k=1
a2kk
�
Mihaly Bencze
PP25474. If xk ∈ R (k = 1, 2, ..., n) ,λk > 0 (k = 1, 2, ..., n) such thatn�
PP25477. In all triangle ABC hold� (r2a+s2)(r2b+s2)
r2a+r2b+2s2≥ 2.
Mihaly Bencze
PP25478. Determine all function f : N∗ → N∗ for whichn�
k=1
k4
f(k) =n4f2(n+1)4f2(n)
for all n ∈ N∗.
Mihaly Bencze
Proposed Problems 321
PP25479. Solve in R the following system:
�4x1+44x2−5
�= 2x3−1
3�4x2+44x3−5
�= 2x4−1
3
−−−−−−−−�4xn+44x1−5
�= 2x2−1
3
, when [·]
denote the integer part.
Mihaly Bencze
PP25480. Solve in R the following system:�x21�− 4 [x2] =
�x22�− 4 [x3] = ... =
�x2n�− 4 [x1] = −3, when [·] denote the
integer part.
Mihaly Bencze
PP25481. Compute S (p, r, n) =n�
k=1
pk�
kr
k+1
�, when [·] denote the integer
part. We have S (2, 2, n) = n · 2n − 2n+1 + 2.
Mihaly Bencze
PP25482. Solve in R the following system
1[x1]
+ 1{x2} = 2016x3
1[x2]
+ 1{x3} = 2016x4
−−−−−−−−−−1
[xn]+ 1
{x1} = 2016x2
, when
[·] and {·} denote the integer respective fractional part.
Mihaly Bencze
PP25483. Determine (an)n≥1 such that a1 = 1 andn�
k=1
ka2k = 14n
2a2n+1 for
all n ≥ 1.
Mihaly Bencze
PP25484. If xk > 0 (k = 1, 2, ..., n) and m ∈ N, then�
cyclic
xm+21 +xm+2
2
xm+11 +xm+1
2
+�
cyclic
xm1 +xm
2
xm−11 +xm−1
2
≤ 2�
cyclic
xm+11 +xm+1
2xm1 +xm
2.
Mihaly Bencze
322 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25485. In all triangle ABC holds� tg2Atg3A+tgA
+� ctg2A
ctg3B+ctgA≥� sin2 2A
1+cos2 2A.
Mihaly Bencze
PP25486. In all triangle ABC holds�(tgA+ ctgA)
�tg2A
tg3A+ctgA+ ctg2A
ctg3A+tgA
�≤ 6.
Mihaly Bencze
PP25487. In all triangle ABC holds 27s2(9t2−1)
≤� 1(st)2−(rarb)
2 ≤ 3t2−2(st)2(t2−1)
for all t > 1.
Mihaly Bencze
PP25488. If x ∈�0, π2�then
�sin2 x
�1 +�
1cos2 x
���+�cos2 x
�1 +�
1sin2 x
���+ 2 =
�1
sin2 x
�+�
1cos2 x
�, when
[·] denote the integer part.
Mihaly Bencze
PP25489. In all triangle ABC holds
3 +�
cyclic
�a
b+c
�1 +�
b+cb+c−a
���=�
cyclic
�b+c
b+c−a
�, when [·] denote the integer
part.
Mihaly Bencze
PP25490. If a1 = 3 then determine (an)n≥1 such thatn�
k=1
(2k−1)(2k+1)a22k−1a
22k+1
= n−13a2n+1
for all n ≥ 2.
Mihaly Bencze
PP25491. If (an)n≥1 is an arithmetical progression with ratio r and a1 ≥ 12 ,
then compute the integer part of the expression�
cyclic
3
�1 + r2
a1a2a3.
Mihaly Bencze
Proposed Problems 323
PP25492. Determine all n, k, r, s ∈ N and all prime p and q such thatn4 + (4k + 1)n2 + 4k2 + 4k + 3 = pr + qs.
Mihaly Bencze
PP25493. Solve in R the following system
x1
�n�
k=1
xk
�= 13
x2
�n�
k=1
xk
�= 23
−−−−−−−−xn
�n�
k=1
xk
�= n3
and prove
thatn�
k=1
3
�x2k =
3
�n2(n+1)2(2n+1)3
108 .
Mihaly Bencze
PP25494. In all triangle ABC holds� tgA
2tgB
2
1+λtgC2
�
tgA2tgB
2
≥ 3λ+3 for all λ > 0.
Mihaly Bencze
PP25495. In all triangle ABC holds���
rarbs2
�2+�rbrcs2
�2+�
s2
rcra
�2≥ 3
√3.
Mihaly Bencze
PP25496. Denote f (n) the number of triples (a, b, c) with properties1 ≤ a ≤ b ≤ c ≤ n when a, b, c are in geometrical progression with ratio a
natural number. Prove that∞�k=1
1f2(k)
> π2
24 .
Mihaly Bencze
PP25497. Solve in N the equation 2015x(x+1) +
2016y(y+1) +
2017z(z+1) = 1.
Mihaly Bencze
PP25498. Compute limx→∞
x
�1
ln 2016 − xlnx
∞�n=0
12016n+x
�
Mihaly Bencze and Jose Luis Dıaz-Barrero
324 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25499. If a, b, c > 0 and a+ b+ c = 1, then 111 +
� a2
5a+2 ≥� (a+b)2
5(a+b)+4
Mihaly Bencze and Jose Luis Dıaz-Barrero
PP25500. If ak > 0 (k = 1, 2, ..., n) andn�
k=1
ak = 1 thenn�
k=1
a2k5ak+2 ≥ 1
2n+5 .
Mihaly Bencze and Jose Luis Dıaz-Barrero
PP25501. In all triangle ABC holds� ab
5b+2(a+b) ≤s6 +
(s2+r2+4Rr)2+8s2Rr
32s(s2+r2+2Rr).
Mihaly Bencze
PP25502. In all triangle ABC holds� 1
ra+ 1
ha
3
�
1+ r2
�
1rb
+ 1hb
�
− r2
�
1rd
+ 1hd
�
≥ 2r .
Mihaly Bencze
PP25503. In all triangle ABC holds 2s2
3s2−r2−4Rr+� w2
abc ≤ 3.
Mihaly Bencze
PP25504. In all triangle ABC holds� (ha)
32
(ha)32√ha+2hb
+� (rb)
32
(ra)32√ra+2rb
≥ 2√r.
Mihaly Bencze
PP25505. In all tetrahedron ABCD holds1).� 3√hbhd
ha3√
hbhd+r(hd−hb)≥ 1
r
2).� 3
√rbrd
ra3√
rbrd+r(rd−rb)≥ 1
r
Mihaly Bencze
PP25506. In all triangle ABC holds� r6a
((4R+r)2+9r2a)2 ≥ (4R+r)2
972 .
Mihaly Bencze
PP25507. In all triangle ABC holds� (s−a)6
(a2+9(s−a)2)2 ≥ s2
972 .
Mihaly Bencze
Proposed Problems 325
PP25508. In all triangle ABC holds
1).� cos A
2cos B
2
cos A2+cos B
2−cos C
2
≥� cos A2
2).� mamb
ma+mb−mc≥�ma
Mihaly Bencze
PP25509. In all triangle ABC holds� c2−(a−b)2
3c−a−b ≥ 2s.
Mihaly Bencze
PP25510. In all triangle ABC holds� (cos A
2+cos B
2 )2
(cos A2+cos B
2 )2−cos2 C
2
≥ 4.
Mihaly Bencze
PP25511. In all triangle ABC holds� a2+b2+3c2+8mamb
4c2−a2−b2+8mamb≥ 4.
Mihaly Bencze
PP25512. In all triangle ABC holds� (a+b)λ
(a+b)λ−cλ≥ 3·2λ
2λ−1for all λ ≥ 1.
Mihaly Bencze
PP25513. Let A1A2...An be a convex polygon. Prove that� (a1+a2+...+an−1)λ
(a1+a2+...+an−1)λ−aλn
≥ n(n−1)λ
(n−1)λ−1for all λ ≥ 1.
Mihaly Bencze
PP25514. If ak, bk, xk > 0 (k = 1, 2, ..., n) and x =n�
k=1
xk then�
n−2n−1
n�k=1
a2k +n�
k=1
a2kxk
x−xk
��n−2n−1
n�k=1
b2k +n�
k=1
b2kxkx−xk
�≥
≥ 4(n−1)2
��
1≤i<j≤n
�aibiajbj
�2
.
Mihaly Bencze
PP25515. If ak > 0, λk > 0 (k = 1, 2, ..., n) andn�
k=1
λk = 1 then
n
�n�
k=1
ak ≤ 1n
�aλ11 aλ2
2 ...aλnn + aλ2
1 aλ32 ...aλ1
n + ...+ aλn1 aλ1
2 ...aλn−1n
�≤ 1
n
n�k=1
ak.
326 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
(A generalization of Heinz-mean).
Mihaly Bencze
PP25516. In all scalene triangle ABC holds� a(b−a)
(b+c)(ma−mb)<
4(2s2−3Rr)3(s2+r2+2Rr)
.
Mihaly Bencze
PP25517. In all scalene triangle ABC holds8Rr(5s2+r2+4Rr)
3(s2+r2+2Rr)+� b3−a3
ma−mb<
2(9s2+r2+4Rr)3 .
Mihaly Bencze
PP25518. In all scalene triangle ABC holds� b2−a2
ma−mb< 16s
3 .
Mihaly Bencze
PP25519. Compute� � (x−sin y cos y−sin2 x)(y−sinx cosx−cos2 y)dxdy
(1−sin 2x)(1−sin 2y) .
Mihaly Bencze
PP25520. In all scalene triangle ABC holds94
� (b−a)mamb
m3a−m3
b≤� ma+mb
a+b ≤ 98
� (b−a)(m2a+m2
b)m3
a−m3b
.
Mihaly Bencze
PP25521. In all acute triangle ABC holds�
(1 + sinA) (1 + cosA) ≥ (3+2√2)sr
R2 .
Mihaly Bencze
PP25522. In all acute triangle ABC holds� 2+3 sinA+2 cos 2A+sinA cos 2A
1+cos 2A ≥ (17+12√2)s
R .
Mihaly Bencze
PP25523. In all acute triangle ABC holds��1 + 1
sinA
� �1 + 1
cosB
�≥ 3(3 + 2
√2).
Mihaly Bencze
Proposed Problems 327
PP25524. Compute� �
�
x+√
y2+y�
(y+√x2+x)dxdy
(x+√x2+x+
√x+
√x+1+1)
�
y+√
y2+y+√y+1+1
� .
Mihaly Bencze
PP25525. Let (an)n≥1 be an arithmetical progression formed with naturalnumbers. Prove that if an thermen of the given arithmetical progression is2025 then the given progression have infinitely many therms formed byperfect square.
Mihaly Bencze
PP25526. If xn+1 · xn = x2n + 1 for all n ≥ 1 and x1 > 1 then compute
limn→∞
n�√
e−�xn+1
xn
�n�.
Mihaly Bencze
PP25527. If xk > 1 (k = 1, 2, ..., n) , then
�cyclic
logx1
�x20142 + x20082 − x20072 + 1
�≥ 2011n.
Mihaly Bencze
PP25528. If xk > 1 (k = 1, 2, ..., n) , then
�cyclic
logx1
�x20142 + x20112 + 2x20082 − x20072 + 1
�≥ 2008n.
Mihaly Bencze
PP25529. If Ak ∈ Mn (C) (k = 1, 2, ..., n) such thatAn−1
1 = −A2A3...An;An−12 = −A3A4...AnA1, ..., A
n−1n = − (A1A2...An−1)
then An1 = An
2 = ... = Ann.
Mihaly Bencze
PP25530. Solve in R the equation
2�√
2x − 1 +√3x − 2x + ...+
�(n+ 1)x − nx
�= (n+ 1)x + n− 1 when
n ∈ N∗.
Mihaly Bencze
328 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25531. Let (an)n≥1 be an arithmetical progression formed by naturalnumbers. Prove that if an term of the given arithmetical progression is 2197the given progression have infinitely many therms formed by perfect cubes.