Generalizations of Popoviciu™s inequality · This criterion entails Vasile Cîrtoaje™s generalization of the Popoviciu inequal- ity (in its standard and in its weighted forms)

Generalizations of Popoviciu�s inequality

Darij Grinberg

25 March 2008

AbstractWe establish a general criterion for inequalities of the kind

convex combination of f (x1) ; f (x2) ; :::; f (xn)

and f (some weighted mean of x1; x2; :::; xn)

� convex combination of f (some other weighted means of x1; x2; :::; xn) ;

where f is a convex function on an interval I � R containing the reals x1; x2;:::; xn; to hold. Here, the left hand side contains only one weighted mean, whilethe right hand side may contain as many as possible, as long as there are �nitelymany. The weighted mean on the left hand side must have positive weights,while those on the right hand side must have nonnegative weights.This criterion entails Vasile Cîrtoaje�s generalization of the Popoviciu inequal-

ity (in its standard and in its weighted forms) as well as a cyclic inequality thatsharpens another result by Vasile Cîrtoaje. The latter cyclic inequality (in itsnon-weighted form) states that

2

nXi=1

f (xi) + n (n� 2) f (x) � nnXs=1

f

�x+

xs � xs+rn

�;

where indices are cyclic modulo n; and x =x1 + x2 + :::+ xn

n:

This is the standard version of this note. A "formal" version with more detailed proofscan be found athttp://www.stud.uni-muenchen.de/~darij.grinberg/PopoviciuFormal.pdf

However, due to these details, it is longer and much more troublesome to read, so it shouldbe used merely as a resort in case you do not understand the proofs in this standard version.

Keywords: Convexity on the real axis, majorization theory, inequalities.

1. Introduction

The last few years saw some activity related to the Popoviciu inequality on convexfunctions. Some generalizations were conjectured and subsequently proven using ma-jorization theory and (mostly) a lot of computations. In this note I am presenting anapparently new approach that proves these generalizations as well as some additionalfacts with a lesser amount of computation and avoiding majorization theory (moreexactly, avoiding the standard, asymmetric de�nition of majorization; we will prove a"symmetric" version of the Karamata inequality on the way, which will not even usethe word "majorize").The very starting point of the whole theory is the following famous fact:

1

Theorem 1a, the Jensen inequality. Let f be a convex function froman interval I � R to R: Let x1; x2; :::; xn be �nitely many points from I:Then,

f (x1) + f (x2) + :::+ f (xn)

n� f

�x1 + x2 + :::+ xn

n

�:

In words, the arithmetic mean of the values of f at the points x1; x2; :::; xnis greater or equal to the value of f at the arithmetic mean of these points.

We can obtain a "weighted version" of this inequality by replacing arithmetic meansby weighted means with some nonnegative weights w1; w2; :::; wn:

Theorem 1b, the weighted Jensen inequality. Let f be a convexfunction from an interval I � R to R: Let x1; x2; :::; xn be �nitely manypoints from I: Let w1; w2; :::; wn be n nonnegative reals which are not allequal to 0: Then,

w1f (x1) + w2f (x2) + :::+ wnf (xn)

w1 + w2 + :::+ wn� f

�w1x1 + w2x2 + :::+ wnxn

w1 + w2 + :::+ wn

�:

Obviously, Theorem 1a follows from Theorem 1b applied to w1 = w2 = ::: = wn = 1;so that Theorem 1b is more general than Theorem 1a.We won�t stop at discussing equality cases here, since they can depend in various

ways on the input (i. e., on the function f; the reals w1; w2; :::; wn and the pointsx1; x2; :::; xn) - but each time we use a result like Theorem 1b, with enough patiencewe can extract the equality case from the proof of this result and the properties of theinput.The Jensen inequality, in both of its versions above, is applied often enough to be

called one of the main methods of proving inequalities. Now, in 1965, a similarly styledinequality was found by the Romanian Tiberiu Popoviciu:

Theorem 2a, the Popoviciu inequality. Let f be a convex functionfrom an interval I � R to R; and let x1; x2; x3 be three points from I:Then,

f (x1)+f (x2)+f (x3)+3f

�x1 + x2 + x3

3

�� 2f

�x2 + x32

�+2f

�x3 + x12

�+2f

�x1 + x22

�:

Again, a weighted version can be constructed:

Theorem 2b, the weighted Popoviciu inequality. Let f be a convexfunction from an interval I � R to R; let x1; x2; x3 be three points fromI; and let w1; w2; w3 be three nonnegative reals such that w2 + w3 6= 0;w3 + w1 6= 0 and w1 + w2 6= 0: Then,

w1f (x1) + w2f (x2) + w3f (x3) + (w1 + w2 + w3) f

�w1x1 + w2x2 + w3x3

w1 + w2 + w3

�� (w2 + w3) f

�w2x2 + w3x3w2 + w3

�+ (w3 + w1) f

�w3x3 + w1x1w3 + w1

�+ (w1 + w2) f

�w1x1 + w2x2w1 + w2

�:

2

The really interesting part of the story began when Vasile Cîrtoaje - alias "Vasc"on the MathLinks forum - proposed the following two generalizations of Theorem 2a([1] and [2] for Theorem 3a, and [1] and [3] for Theorem 4a):

Theorem 3a (Vasile Cîrtoaje). Let f be a convex function from aninterval I � R to R: Let x1; x2; :::; xn be �nitely many points from I:Then,

nXi=1

f (xi)+n (n� 2) f�x1 + x2 + :::+ xn

n

��

nXj=1

(n� 1) f

0B@P

1�i�n; i6=jxi

n� 1

1CA :

Theorem 4a (Vasile Cîrtoaje). Let f be a convex function from aninterval I � R to R: Let x1; x2; :::; xn be �nitely many points from I:Then,

(n� 2)nXi=1

f (xi) + nf

�x1 + x2 + :::+ xn

n

��

X1�i<j�n

2f

�xi + xj2

�:

In [1], both of these facts were nicely proven by Cîrtoaje. I gave a di¤erent andrather long proof of Theorem 3a in [2]. All of these proofs use the Karamata inequality.Theorem 2a follows from each of the Theorems 3a and 4a upon setting n = 3:It is pretty straightforward to obtain generalizations of Theorems 3a and 4a by

putting in weights as in Theorems 1b and 2b. A more substantial generalization wasgiven by Yufei Zhao - alias "Billzhao" on MathLinks - in [3]:

Theorem 5a (Yufei Zhao). Let f be a convex function from an intervalI � R to R: Let x1; x2; :::; xn be �nitely many points from I; and let m bean integer. Then,�

n� 2m� 1

� nXi=1

f (xi) +

�n� 2m� 2

�nf

�x1 + x2 + :::+ xn

n

��

X1�i1<i2<:::<im�n

mf

�xi1 + xi2 + :::+ xim

m

�:

Note that if m � 0 or m > n; the sumP

1�i1<i2<:::<im�nmf

�xi1 + xi2 + :::+ xim

m

�is

empty, so that its value is 0:Note that Theorems 3a and 4a both are particular cases of Theorem 5a (in fact, set

m = n� 1 to get Theorem 3a and m = 2 to get Theorem 4a).An rather complicated proof of Theorem 5a was given by myself in [3]. After some

time, the MathLinks user "Zhaobin" proposed a weighted version of this result:

Theorem 5b (Zhaobin). Let f be a convex function from an intervalI � R to R: Let x1; x2; :::; xn be �nitely many points from I; let w1;w2; :::; wn be nonnegative reals, and let m be an integer. Assume that

3

w1+w2+ :::+wn 6= 0; and that wi1 +wi2 + :::+wim 6= 0 for any m integersi1; i2; :::; im satisfying 1 � i1 < i2 < ::: < im � n:

Then,�n� 2m� 1

� nXi=1

wif (xi) +

�n� 2m� 2

�(w1 + w2 + :::+ wn) f

�w1x1 + w2x2 + :::+ wnxn

w1 + w2 + :::+ wn

��

X1�i1<i2<:::<im�n

(wi1 + wi2 + :::+ wim) f

�wi1xi1 + wi2xi2 + :::+ wimxim

wi1 + wi2 + :::+ wim

�:

If we set w1 = w2 = ::: = wn = 1 in Theorem 5b, we obtain Theorem 5a. On theother hand, putting n = 3 and m = 2 in Theorem 5b, we get Theorem 2b.In this note, I am going to prove Theorem 5b (and therefore also its particular cases

- Theorems 2a, 2b, 3a, 4a and 5a). The proof is going to use no preknowledge - inparticular, classical majorization theory will be avoided. Then, we are going to discussan assertion similar to Theorem 5b with its applications.

2. Absolute values interpolate convex functions

We start preparing for our proof by showing a classical property of convex func-tions1:

Theorem 6 (Hardy, Littlewood, Pólya). Let f be a convex functionfrom an interval I � R to R: Let x1; x2; :::; xn be �nitely many points fromI: Then, there exist two real constants u and v and n nonnegative constantsa1; a2; :::; an such that

f (t) = vt+ u+nXi=1

ai jt� xij holds for every t 2 fx1; x2; :::; xng :

In brief, this result states that every convex function f (x) on n reals x1; x2; :::; xncan be interpolated by a linear combination with nonnegative coe¢ cients of a linearfunction and the n functions jx� xij :The proof of Theorem 6, albeit technical, will be given here for the sake of complete-

ness: First, we need an almost trivial fact which I use to call the max f0; xg formula:For any real number x; we have max f0; xg = 1

2(x+ jxj) :

Furthermore, we denote f [y; z] =f (y)� f (z)

y � zfor any two points y and z from I

satisfying y 6= z: Then, we have (y � z) � f [y; z] = f (y) � f (z) for any two points yand z from I satisfying y 6= z:We can assume that all points x1; x2; :::; xn are pairwisely distinct (if not, we can

remove all super�uous xi and apply Theorem 6 to the remaining points). Therefore,

1This property appeared as Proposition B.4 in [8], which refers to [9] for its origins. It was alsomentioned by a MathLinks user called "Fleeting_Guest" in [4], post #18 as a known fact, albeit ina slightly di¤erent (but equivalent) form.

4

we can WLOG assume that x1 < x2 < ::: < xn: Then, for every j 2 f1; 2; :::; ng ; wehave

f (xj) = f (x1) +

j�1Xk=1

(f (xk+1)� f (xk)) = f (x1) +

j�1Xk=1

(xk+1 � xk) � f [xk+1; xk]

= f (x1) +

j�1Xk=1

(xk+1 � xk) � f [x2; x1] +

kXi=2

(f [xi+1; xi]� f [xi; xi�1])

!

= f (x1) +

j�1Xk=1

(xk+1 � xk) � f [x2; x1] +j�1Xk=1

(xk+1 � xk) �kXi=2

(f [xi+1; xi]� f [xi; xi�1])

= f (x1) + f [x2; x1] �j�1Xk=1

(xk+1 � xk) +

j�1Xk=1

kXi=2

(f [xi+1; xi]� f [xi; xi�1]) � (xk+1 � xk)

= f (x1) + f [x2; x1] �j�1Xk=1

(xk+1 � xk) +

j�1Xi=2

j�1Xk=i

(f [xi+1; xi]� f [xi; xi�1]) � (xk+1 � xk)

= f (x1) + f [x2; x1] �j�1Xk=1

(xk+1 � xk) +

j�1Xi=2

(f [xi+1; xi]� f [xi; xi�1]) �j�1Xk=i

(xk+1 � xk)

= f (x1) + f [x2; x1] � (xj � x1) +

j�1Xi=2

(f [xi+1; xi]� f [xi; xi�1]) � (xj � xi) :

Now we set

�1 = �n = 0;

�i = f [xi+1; xi]� f [xi; xi�1] for all i 2 f2; 3; :::; n� 1g :

5

Using these notations, the above computation becomes

f (xj) = f (x1) + f [x2; x1] � (xj � x1) +

j�1Xi=2

�i � (xj � xi)

= f (x1) + f [x2; x1] � (xj � x1) + 0|{z}=�1

�max f0; xj � x1g

+

j�1Xi=2

�i � (xj � xi)| {z }=maxf0;xj�xig; since xj�xi�0; as xi�xj

+

nXi=j

�i � 0|{z}=maxf0;xj�xig; since xj�xi�0; as xj�xi

= f (x1) + f [x2; x1] � (xj � x1) + �1 �max f0; xj � x1g

+

j�1Xi=2

�i �max f0; xj � xig+nXi=j

�i �max f0; xj � xig

= f (x1) + f [x2; x1] � (xj � x1) +nXi=1

�i �max f0; xj � xig

= f (x1) + f [x2; x1] � (xj � x1) +nXi=1

�i �1

2((xj � xi) + jxj � xij)�

since max f0; xj � xig =1

2((xj � xi) + jxj � xij) by the max f0; xg formula

�= f (x1) + f [x2; x1] � (xj � x1) +

nXi=1

�i �1

2(xj � xi) +

nXi=1

�i �1

2jxj � xij

= f (x1) + (f [x2; x1]xj � f [x2; x1]x1) +

1

2

nXi=1

�ixj �1

2

nXi=1

�ixi

!+

nXi=1

1

2�i jxj � xij

=

f [x2; x1] +

1

2

nXi=1

�i

!xj +

f (x1)� f [x2; x1]x1 �

1

2

nXi=1

�ixi

!+

nXi=1

1

2�i jxj � xij :

Thus, if we denote

v = f [x2; x1] +1

2

nXi=1

�i; u = f (x1)� f [x2; x1]x1 �1

2

nXi=1

�ixi;

ai =1

2�i for all i 2 f1; 2; :::; ng ;

then we have

f (xj) = vxj + u+nXi=1

ai jxj � xij :

Since we have shown this for every j 2 f1; 2; :::; ng ; we can restate this as follows: Wehave

f (t) = vt+ u+

nXi=1

ai jt� xij for every t 2 fx1; x2; :::; xng :

Hence, in order for the proof of Theorem 6 to be complete, it is enough to show that

the n reals a1; a2; :::; an are nonnegative. Since ai =1

2�i for every i 2 f1; 2; :::; ng ;

6

this will follow once it is proven that the n reals �1; �2; :::; �n are nonnegative. Thus,we have to show that �i is nonnegative for every i 2 f1; 2; :::; ng : This is trivial fori = 1 and for i = n (since �1 = 0 and �n = 0), so it remains to prove that �i isnonnegative for every i 2 f2; 3; :::; n� 1g : Now, since �i = f [xi+1; xi] � f [xi; xi�1]for every i 2 f2; 3; :::; n� 1g ; we thus have to show that f [xi+1; xi] � f [xi; xi�1] isnonnegative for every i 2 f2; 3; :::; n� 1g : In other words, we have to prove thatf [xi+1; xi] � f [xi; xi�1] for every i 2 f2; 3; :::; n� 1g : But since xi�1 < xi < xi+1; thisfollows from the next lemma:

Lemma 7. Let f be a convex function from an interval I � R to R: Let x;y; z be three points from I satisfying x < y < z: Then, f [z; y] � f [y; x] :

Proof of Lemma 7. Since the function f is convex on I; and since z and x are pointsfrom I; the de�nition of convexity yields

1

z � yf (z) +

1

y � xf (x)

1

z � y+

1

y � x

� f

0BB@1

z � yz +

1

y � xx

1

z � y+

1

y � x

1CCA(here we have used that

1

z � y> 0 and

1

y � x> 0; what is clear from x < y < z).

Since

1

z � yz +

1

y � xx

1

z � y+

1

y � x

= y; this simpli�es to

1

z � yf (z) +

1

y � xf (x)

1

z � y+

1

y � x

� f (y) ; so that

1

z � yf (z) +

1

y � xf (x) �

�1

z � y+

1

y � x

�f (y) ; so that

1

z � yf (z)� 1

z � yf (y) � 1

y � xf (y)� 1

y � xf (x) ; so that

f (z)� f (y)

z � y� f (y)� f (x)

y � x:

This becomes f [z; y] � f [y; x] ; and thus Lemma 7 is proven. Thus, the proof ofTheorem 6 is completed.

3. The Karamata inequality in symmetric form

Now as Theorem 6 is proven, it becomes easy to prove the Karamata inequality inthe following form:

Theorem 8a, the Karamata inequality in symmetric form. Let fbe a convex function from an interval I � R to R; and let n be a positiveinteger. Let x1; x2; :::; xn; y1; y2; :::; yn be 2n points from I: Assume that

jx1 � tj+ jx2 � tj+ :::+ jxn � tj � jy1 � tj+ jy2 � tj+ :::+ jyn � tj

7

holds for every t 2 fx1; x2; :::; xn; y1; y2; :::; yng : Then,

f (x1) + f (x2) + :::+ f (xn) � f (y1) + f (y2) + :::+ f (yn) :

We will not need this result, but we will rather use its weighted version:

Theorem 8b, the weighted Karamata inequality in symmetricform. Let f be a convex function from an interval I � R to R; andlet N be a positive integer. Let z1; z2; :::; zN be N points from I; and letw1; w2; :::; wN be N reals. Assume that

NXk=1

wk = 0; (1)

and that

NXk=1

wk jzk � tj � 0 holds for every t 2 fz1; z2; :::; zNg : (2)

Then,NXk=1

wkf (zk) � 0: (3)

It is very easy to conclude Theorem 8a from Theorem 8b by setting N = 2n and

z1 = x1; z2 = x2; :::; zn = xn;

zn+1 = y1; zn+2 = y2; :::; z2n = yn;

w1 = w2 = ::: = wn = 1; wn+1 = wn+2 = ::: = w2n = �1;

but as I said, we will never use Theorem 8a in this paper.Time for a remark to readers familiar with majorization theory. One may wonder

why I call the two results above "Karamata inequalities". In fact, the Karamatainequality in its most known form claims:

Theorem 9, the Karamata inequality. Let f be a convex function froman interval I � R to R; and let n be a positive integer. Let x1; x2; :::; xn;y1; y2; :::; yn be 2n points from I such that (x1; x2; :::; xn) � (y1; y2; :::; yn) :Then,

f (x1) + f (x2) + :::+ f (xn) � f (y1) + f (y2) + :::+ f (yn) :

According to [2], post #11, Lemma 1, the condition (x1; x2; :::; xn) � (y1; y2; :::; yn)yields that jx1 � tj + jx2 � tj + ::: + jxn � tj � jy1 � tj + jy2 � tj + ::: + jyn � tj holdsfor every real t - and thus, in particular, for every t 2 fz1; z2; :::; zng : Hence, wheneverthe condition of Theorem 9 holds, the condition of Theorem 8a holds as well. Thus,Theorem 9 follows from Theorem 8a. With just a little more work, we could also deriveTheorem 8a from Theorem 9, so that Theorems 8a and 9 are equivalent.

8

Note that Theorem 8b is more general than the Fuchs inequality (a more well-known weighted version of the Karamata inequality). See [5] for a generalization ofmajorization theory to weighted families of points (apparently already known long timeago), with a di¤erent approach to this fact.As promised, here is a proof of Theorem 8b: First, substituting t = max fz1; z2; :::; zNg

into (2) (it is clear that this t satis�es t 2 fz1; z2; :::; zNg), we get zk � t for everyk 2 f1; 2; :::; Ng ; so that zk � t � 0 and thus jzk � tj = � (zk � t) = t � zk for

every k 2 f1; 2; :::; Ng ; and thus (2) becomesNPk=1

wk (t� zk) � 0: In other words,

tNPk=1

wk �NPk=1

wkzk � 0: In sight ofNPk=1

wk = 0; this rewrites as t � 0 �NPk=1

wkzk � 0:

Hence,NPk=1

wkzk � 0:

Similarly, substituting t = min fz1; z2; :::; zNg into (2), we getNPk=1

wkzk � 0: Thus,NPk=1

wkzk = 0:

The function f : I ! R is convex, and z1; z2; :::; zN are �nitely many points from I:Hence, Theorem 6 yields the existence of two real constants u and v and N nonnegativeconstants a1; a2; :::; aN such that

f (t) = vt+ u+NXi=1

ai jt� zij holds for every t 2 fz1; z2; :::; zNg :

Thus,

f (zk) = vzk + u+NXi=1

ai jzk � zij for every k 2 f1; 2; :::; Ng :

Hence,

NXk=1

wkf (zk) =NXk=1

wk

vzk + u+

NXi=1

ai jzk � zij!= v

NXk=1

wkzk| {z }=0

+u

NXk=1

wk| {z }=0

+

NXk=1

wk

NXi=1

ai jzk � zij

=

NXk=1

wk

NXi=1

ai jzk � zij =NXi=1

ai

NXk=1

wk jzk � zij| {z }�0 according to (2) for t=zi

� 0:

Thus, Theorem 8b is proven.

4. A property of zero-sum vectors

Next, we are going to show some properties of real vectors.If k is an integer and v 2 Rk is a vector, then, for any i 2 f1; 2; :::; kg ; we denote

by vi the i-th coordinate of the vector v: Then, v =

0BB@v1v2:::vk

1CCA :

9

Let n be a positive integer. We consider the vector space Rn: Let (e1; e2; :::; en) bethe standard basis of this vector space Rn; in other words, for every i 2 f1; 2; :::; ng ; letei be the vector from Rn such that (ei)i = 1 and (ei)j = 0 for every j 2 f1; 2; :::; ngnfig :Let Vn be the subspace of Rn de�ned by

Vn = fx 2 Rn j x1 + x2 + :::+ xn = 0g :

For any u 2 f1; 2; :::; ng and any two distinct numbers i and j from the setf1; 2; :::; ng ; we have

(ei � ej)u =

8<:1; if u = i;�1; if u = j;

0; if u 6= i and u 6= j: (4)

Clearly, ei � ej 2 Vn for any two numbers i and j from the set f1; 2; :::; ng :For any vector t 2 Rn; we denote I (t) = fk 2 f1; 2; :::; ng j tk > 0g and J (t) =

fk 2 f1; 2; :::; ng j tk < 0g : Obviously, for every t 2 Rn; the sets I (t) and J (t) aredisjoint.Now we are going to show:

Theorem 10. Let n be a positive integer. Let x 2 Vn be a vector. Then,there exist nonnegative reals ai;j for all pairs (i; j) 2 I (x)�J (x) such that

x =X

(i;j)2I(x)�J(x)

ai;j (ei � ej) :

Proof of Theorem 10. We will prove Theorem 10 by induction over jI (x)j+ jJ (x)j :The basis of the induction - the case when jI (x)j+ jJ (x)j = 0 - is trivial: If jI (x)j+

jJ (x)j = 0; then I (x) = J (x) = ? and x = 0; so that x =P

(i;j)2I(x)�J(x)ai;j (ei � ej)

holds because both sides of this equation are 0:Now we come to the induction step: Let r be a positive integer. Assume that

Theorem 10 holds for all x 2 Vn with jI (x)j + jJ (x)j < r: We have to show thatTheorem 10 holds for all x 2 Vn with jI (x)j+ jJ (x)j = r:In order to prove this, we let z 2 Vn be an arbitrary vector with jI (z)j+ jJ (z)j = r:

We then have to prove that Theorem 10 holds for x = z: In other words, we have toshow that there exist nonnegative reals ai;j for all pairs (i; j) 2 I (z)� J (z) such that

z =X

(i;j)2I(z)�J(z)

ai;j (ei � ej) : (5)

First, jI (z)j + jJ (z)j = r and r > 0 yield jI (z)j + jJ (z)j > 0: Hence, at least oneof the sets I (z) and J (z) is non-empty.Now, since z 2 Vn; we have z1 + z2 + ::: + zn = 0: Hence, either zk = 0 for every

k 2 f1; 2; :::; ng ; or there is at least one positive number and at least one negativenumber in the set fz1; z2; :::; zng : The �rst case is impossible (since at least one of thesets I (z) and J (z) is non-empty). Thus, the second case must hold - i. e., there is atleast one positive number and at least one negative number in the set fz1; z2; :::; zng :In other words, there exists a number u 2 f1; 2; :::; ng such that zu > 0; and a number

10

v 2 f1; 2; :::; ng such that zv < 0: Of course, zu > 0 yields u 2 I (z) ; and zv < 0 yieldsv 2 J (z) : Needless to say that u 6= v:Now, we distinguish between two cases: the �rst case will be the case when zu+zv �

0; and the second case will be the case when zu + zv � 0:Let us consider the �rst case: In this case, zu+zv � 0: Then, let z0 = z+zv (eu � ev) :

Since z 2 Vn and eu�ev 2 Vn; we have z+zv (eu � ev) 2 Vn (since Vn is a vector space),so that z0 2 Vn: From z0 = z+ zv (eu � ev) ; the coordinate representation of the vectorz0 is easily obtained:

z0 =

0BB@z01z02:::z0n

1CCA ; where

8<:z0k = zk for all k 2 f1; 2; :::; ng n fu; vg ;

z0u = zu + zv;z0v = 0

:

It is readily seen from this that I (z0) � I (z), so that jI (z0)j � jI (z)j : Besides, J (z0) �J (z) : Moreover, J (z0) is a proper subset of J (z) ; because v =2 J (z0) (since z0v is not< 0; but = 0) but v 2 J (z) : Hence, jJ (z0)j < jJ (z)j : Combined with jI (z0)j � jI (z)j ;this yields jI (z0)j + jJ (z0)j < jI (z)j + jJ (z)j : In view of jI (z)j + jJ (z)j = r; thisbecomes jI (z0)j+ jJ (z0)j < r: Thus, since we have assumed that Theorem 10 holds forall x 2 Vn with jI (x)j + jJ (x)j < r; we can apply Theorem 10 to x = z0; and we seethat there exist nonnegative reals a0i;j for all pairs (i; j) 2 I (z0)� J (z0) such that

z0 =X

(i;j)2I(z0)�J(z0)

a0i;j (ei � ej) :

Now, z0 = z + zv (eu � ev) yields z = z0 � zv (eu � ev) : Since zv < 0; we have �zv > 0;so that, particularly, �zv is nonnegative.Since I (z0) � I (z) and J (z0) � J (z) ; we have I (z0)� J (z0) � I (z)� J (z) : Also,

(u; v) 2 I (z) � J (z) (because u 2 I (z) and v 2 J (z)) and (u; v) =2 I (z0) � J (z0)(because v =2 J (z0)).Hence, the sets I (z0)�J (z0) and f(u; v)g are two disjoint subsets of the set I (z)�

J (z) : We can thus de�ne nonnegative reals ai;j for all pairs (i; j) 2 I (z) � J (z) bysetting

ai;j =

8<:a0i;j; if (i; j) 2 I (z0)� J (z0) ;

�zv; if (i; j) = (u; v) ;0; if neither of the two cases above holds

(these ai;j are all nonnegative because a0i;j; �zv and 0 are nonnegative). Then,X(i;j)2I(z)�J(z)

ai;j (ei � ej)

=X

(i;j)2I(z0)�J(z0)

a0i;j (ei � ej) +X

(i;j)=(u;v)

(�zv) (ei � ej) + (sum of some 0�s)

=X

(i;j)2I(z0)�J(z0)

a0i;j (ei � ej) + (�zv) (eu � ev) + 0 = z0 + (�zv) (eu � ev) + 0

= (z + zv (eu � ev)) + (�zv) (eu � ev) + 0 = z:

Thus, (5) is ful�lled.

11

Similarly, we can ful�ll (5) in the second case, repeating the arguments we havedone for the �rst case while occasionally interchanging u with v; as well as I with J; aswell as < with >. Here is a brief outline of how we have to proceed in the second case:Denote z0 = z � zu (eu � ev) : Show that z0 2 Vn (as in the �rst case). Notice that

z0 =

0BB@z01z02:::z0n

1CCA ; where

8<:z0k = zk for all k 2 f1; 2; :::; ng n fu; vg ;

z0u = 0;z0v = zu + zv

:

Prove that u =2 I (z0) (as we proved v =2 J (z0) in the �rst case). Prove that J (z0) � J (z)(similarly to the proof of I (z0) � I (z) in the �rst case) and that I (z0) is a proper subsetof I (z) (similarly to the proof that J (z0) is a proper subset of J (z) in the �rst case).Show that there exist nonnegative reals a0i;j for all pairs (i; j) 2 I (z0)�J (z0) such that

z0 =X

(i;j)2I(z0)�J(z0)

a0i;j (ei � ej)

(as in the �rst case). Note that zu is nonnegative (since zu > 0). Prove that the setsI (z0) � J (z0) and f(u; v)g are two disjoint subsets of the set I (z) � J (z) (as in the�rst case). De�ne nonnegative reals ai;j for all pairs (i; j) 2 I (z)� J (z) by setting

ai;j =

8<:a0i;j; if (i; j) 2 I (z0)� J (z0) ;

zu; if (i; j) = (u; v) ;0; if neither of the two cases above holds

:

Prove that these nonnegative reals ai;j ful�ll (5).Thus, in each of the two cases, we have proven that there exist nonnegative reals

ai;j for all pairs (i; j) 2 I (z) � J (z) such that (5) holds. Hence, Theorem 10 holdsfor x = z: Thus, Theorem 10 is proven for all x 2 Vn with jI (x)j + jJ (x)j = r: Thiscompletes the induction step, and therefore, Theorem 10 is proven.As an application of Theorem 10, we can now show:

Theorem 11. Let n be a positive integer. Let a1; a2; :::; an be n nonneg-ative reals. Let S be a �nite set. For every s 2 S; let rs be an element of(Rn)� (in other words, a linear transformation from Rn to R), and let bs bea nonnegative real. De�ne a function f : Rn ! R by

f (x) =

nXu=1

au jxuj �Xs2S

bs jrsxj ; where x =

0BB@x1x2:::xn

1CCA 2 Rn:

Then, the following two assertions are equivalent:

Assertion A1: We have f (x) � 0 for every x 2 Vn:Assertion A2: We have f (ei � ej) � 0 for any two distinct integers i andj from f1; 2; :::; ng :

12

Proof of Theorem 11. We have to prove that the assertionsA1 andA2 are equivalent.In other words, we have to prove that A1 =) A2 and A2 =) A1: Actually, A1 =) A2is trivial (we just have to use that ei � ej 2 Vn for any two numbers i and j fromf1; 2; :::; ng). It remains to show that A2 =) A1: So assume that Assertion A2 is valid,i. e. we have f (ei � ej) � 0 for any two distinct integers i and j from f1; 2; :::; ng :Wehave to prove that Assertion A1 holds, i. e. that f (x) � 0 for every x 2 Vn:So let x 2 Vn be some vector. According to Theorem 10, there exist nonnegative

reals ai;j for all pairs (i; j) 2 I (x)� J (x) such that

x =X

(i;j)2I(x)�J(x)

ai;j (ei � ej) :

We will now show that

jxuj =X

(i;j)2I(x)�J(x)

ai;j��(ei � ej)u

�� for every u 2 f1; 2; :::; ng : (6)

Here, of course, (ei � ej)u means the u-th coordinate of the vector ei � ej:In fact, two cases are possible: the case when xu � 0; and the case when xu < 0:

We will consider these cases separately.Case 1: We have xu � 0: Then, jxuj = xu: Hence, in this case, we have (ei � ej)u � 0

for any two numbers i 2 I (x) and j 2 J (x) (in fact, j 2 J (x) yields xj < 0; sothat u 6= j (because xj < 0 and xu � 0) and thus (ej)u = 0; so that (ei � ej)u =

(ei)u � (ej)u = (ei)u � 0 = (ei)u =�1; if u = i;0; if u 6= i

� 0). Thus, (ei � ej)u =��(ei � ej)u

��for any two numbers i 2 I (x) and j 2 J (x) : Thus,

jxuj = xu =X

(i;j)2I(x)�J(x)

ai;j (ei � ej)u

0@since x = X(i;j)2I(x)�J(x)

ai;j (ei � ej)

1A=

X(i;j)2I(x)�J(x)


�� ;and (6) is proven.Case 2: We have xu < 0: Then, u 2 J (x) and jxuj = �xu: Hence, in this case,

we have (ei � ej)u � 0 for any two numbers i 2 I (x) and j 2 J (x) (in fact, i 2 I (x)yields xi > 0; so that u 6= i (because xi > 0 and xu < 0) and thus (ei)u = 0; so

that (ei � ej)u = (ei)u � (ej)u = 0 � (ej)u = � (ej)u = ��1; if u = j;0; if u 6= j

� 0). Thus,

� (ei � ej)u =��(ei � ej)u

�� for any two numbers i 2 I (x) and j 2 J (x) : Thus,jxuj = �xu = �

X(i;j)2I(x)�J(x)

ai;j (ei � ej)u

0@since x = X(i;j)2I(x)�J(x)

ai;j (ei � ej)

1A=

X(i;j)2I(x)�J(x)

ai;j�� (ei � ej)u

�=

X(i;j)2I(x)�J(x)


�� ;and (6) is proven.

13

Hence, in both cases, (6) is proven. Thus, (6) always holds. Now let us continueour proof of A2 =) A1:We haveX

s2Sbs jrsxj =

Xs2S

bs

��rsX

(i;j)2I(x)�J(x)

ai;j (ei � ej)

��0@since x = X

(i;j)2I(x)�J(x)

ai;j (ei � ej)

1A=Xs2S

bs

��X

(i;j)2I(x)�J(x)

ai;jrs (ei � ej)

��Xs2S

bsX

(i;j)2I(x)�J(x)

ai;j jrs (ei � ej)j

(by the triangle inequality, since all ai;j and all bs are nonnegative) :

Thus,

f (x) =nXu=1

au jxuj �Xs2S

bs jrsxj �nXu=1

au jxuj �Xs2S

bsX

(i;j)2I(x)�J(x)

ai;j jrs (ei � ej)j

=nXu=1

au �X

(i;j)2I(x)�J(x)


��Xs2S

bsX

(i;j)2I(x)�J(x)

ai;j jrs (ei � ej)j (by (6))

=X

(i;j)2I(x)�J(x)

ai;j

nXu=1

au��(ei � ej)u

�� X(i;j)2I(x)�J(x)

ai;jXs2S

bs jrs (ei � ej)j

=X

(i;j)2I(x)�J(x)

ai;j �

nXu=1


��Xs2S

bs jrs (ei � ej)j!

=X

(i;j)2I(x)�J(x)

ai;j|{z}�0

� f (ei � ej)| {z }�0

� 0:

(Here, f (ei � ej) � 0 because i and j are two distinct integers from f1; 2; :::; ng ; infact, i and j are distinct because i 2 I (x) and j 2 J (x) ; and the sets I (x) and J (x)are disjoint.)Hence, we have obtained f (x) � 0: This proves the assertion A1: Therefore, the

implication A2 =) A1 is proven, and the proof of Theorem 11 is complete.

5. Restating Theorem 11

Now we consider a result which follows from Theorem 11 pretty obviously (althoughthe formalization of the proof is going to be gruelling):

Theorem 12. Let n be a nonnegative integer. Let a1; a2; :::; an and a ben+1 nonnegative reals. Let S be a �nite set. For every s 2 S; let rs be anelement of (Rn)� (in other words, a linear transformation from Rn to R),and let bs be a nonnegative real. De�ne a function g : Rn ! R by

g (x) =

nXu=1

au jxuj+a jx1 + x2 + :::+ xnj�Xs2S


0BB@x1x2:::xn

1CCA 2 Rn:

14


Assertion B1: We have g (x) � 0 for every x 2 Rn:Assertion B2: We have g (ei) � 0 for every integer i 2 f1; 2; :::; ng ; andg (ei � ej) � 0 for any two distinct integers i and j from f1; 2; :::; ng :

Proof of Theorem 12. We are going to restate Theorem 12 before we actually proveit. But �rst, we introduce a notation:Let (ee1; ee2; :::; gen�1) be the standard basis of the vector space Rn�1; in other words,

for every i 2 f1; 2; :::; n� 1g ; let eei be the vector from Rn�1 such that (eei)i = 1 and(eei)j = 0 for every j 2 f1; 2; :::; n� 1g n fig :Now we will restate Theorem 12 by renaming n into n� 1 (thus replacing ei by eei

as well) and a into an:

Theorem 12b. Let n be a positive integer. Let a1; a2; :::; an�1; an ben nonnegative reals. Let S be a �nite set. For every s 2 S; let rs be anelement of (Rn�1)� (in other words, a linear transformation from Rn�1 toR), and let bs be a nonnegative real. De�ne a function g : Rn�1 ! R by

g (x) =n�1Xu=1

au jxuj+an jx1 + x2 + :::+ xn�1j�Xs2S


0BB@x1x2:::xn�1

1CCA 2 Rn�1:


Assertion C1: We have g (x) � 0 for every x 2 Rn�1:Assertion C2: We have g (eei) � 0 for every integer i 2 f1; 2; :::; n� 1g ; andg (eei � eej) � 0 for any two distinct integers i and j from f1; 2; :::; n� 1g :

Theorem 12b is equivalent to Theorem 12 (because Theorem 12b is just Theorem12, applied to n� 1 instead of n). Thus, proving Theorem 12b will be enough to verifyTheorem 12.Proof of Theorem 12b. The implication C1 =) C2 is absolutely trivial. Hence, in

order to establish Theorem 12b, it only remains to prove the implication C2 =) C1:So assume that the assertion C2 holds, i. e. that we have g (eei) � 0 for every

integer i 2 f1; 2; :::; n� 1g ; and g (eei � eej) � 0 for any two distinct integers i and jfrom f1; 2; :::; n� 1g : We want to show that Assertion C1 holds, i. e. that g (x) � 0 issatis�ed for every x 2 Rn�1:Since (ee1; ee2; :::; gen�1) is the standard basis of the vector space Rn�1; every vector

x 2 Rn�1 satis�es x =n�1Pi=1

xieei:Since (e1; e2; :::; en) is the standard basis of the vector space Rn; every vector x 2 Rn

satis�es x =nPi=1

xiei:

Let �n : Rn�1 ! Rn be the linear transformation de�ned by �neei = ei�en for everyi 2 f1; 2; :::; n� 1g : (This linear transformation is uniquely de�ned this way because

15

(ee1; ee2; :::; gen�1) is a basis of Rn�1:) For every x 2 Rn�1; we then have�nx = �n

n�1Xi=1

xieei! = n�1Xi=1

xi�neei (since �n is linear)

=

n�1Xi=1

xi (ei � en) =

n�1Xi=1

xiei �n�1Xi=1

xien =

n�1Xi=1

xiei � (x1 + x2 + :::+ xn�1) en

=

0BBBB@x1x2:::xn�1

� (x1 + x2 + :::+ xn�1)

1CCCCA ; (7)

Consequently, �nx 2 Vn for every x 2 Rn�1. Hence, Im�n � Vn:

Let n : Rn ! Rn�1 be the linear transformation de�ned by nei =� eei; if i 2 f1; 2; :::; n� 1g ;

0; if i = nfor every i 2 f1; 2; :::; ng : (This linear transformation is uniquely de�ned this way be-cause (e1; e2; :::; en) is a basis of Rn:) For every x 2 Rn; we then have

nx = n

nXi=1

xiei

!=

nXi=1

xi nei (since n is linear)

=nXi=1

xi

� eei; if i 2 f1; 2; :::; n� 1g ;0; if i = n

=n�1Xi=1

xieei =0BB@

x1x2:::xn�1

1CCA :

Then, n�n = id (in fact, for every i 2 f1; 2; :::; n� 1g ; we have

n�neei = n (ei � en) = nei � nen (since n is linear)

= eei � 0 = eei;thus, for every x 2 Rn�1; we have

n�nx = n�n

n�1Xi=1

xieei! = n�1Xi=1

xi n�neei(since the function n�n is linear, because n and �n are linear)

=

n�1Xi=1

xieei = x;

and therefore n�n = id).We de�ne a function f : Rn ! R by

f (x) =nXu=1

au jxuj �Xs2S

bs jrs nxj ; where x =

0BB@x1x2:::xn

1CCA 2 Rn:

16

Note thatf (�x) = f (x) for every x 2 Rn; (8)

since

f (�x) =nXu=1

au j(�x)uj �Xs2S

bs jrs n (�x)j =nXu=1

au j�xuj �Xs2S

bs j�rs nxj

(here, we have rs n (�x) = �rs nx since rs and n are linear functions)

=nXu=1

au jxuj �Xs2S

bs jrs nxj = f (x) :

Furthermore, I claim that

f (�nx) = g (x) for every x 2 Rn�1: (9)

In order to prove this, we note that (7) yields (�nx)u = xu for all u 2 f1; 2; :::; n� 1gand (�nx)n = � (x1 + x2 + :::+ xn�1) ; while n�n = id yields n�nx = x; so that

f (�nx) =nXu=1

au j(�nx)uj �Xs2S

bs jrs n�nxj

=n�1Xu=1

au j(�nx)uj+ an j(�nx)nj �Xs2S

bs jrs n�nxj

=n�1Xu=1

au jxuj+ an j� (x1 + x2 + :::+ xn�1)j �Xs2S

bs jrsxj

(since (�nx)u = xu for all u 2 f1; 2; :::; n� 1g and(�nx)n = � (x1 + x2 + :::+ xn�1) ; and n�nx = x)

=n�1Xu=1

au jxuj+ an jx1 + x2 + :::+ xn�1j �Xs2S

bs jrsxj = g (x) ;

and thus (9) is proven.Now, we are going to show that

f (ei � ej) � 0 for any two distinct integers i and j from f1; 2; :::; ng : (10)

In order to prove (10), we distinguish between three di¤erent cases:Case 1: We have i 2 f1; 2; :::; n� 1g and j 2 f1; 2; :::; n� 1g :Case 2: We have i 2 f1; 2; :::; n� 1g and j = n:Case 3: We have i = n and j 2 f1; 2; :::; n� 1g :(In fact, the case when both i = n and j = n cannot occur, since i and j must be

distinct).In Case 1, we have

f (ei � ej) = f ((ei � en)� (ej � en)) = f (�neei � �n eej)= f (�n (eei � eej)) (since �neei � �n eej = �n (eei � eej) ; because �n is linear)= g (eei � eej) (after (9))

� 0 (by assumption) :

17

In Case 2, we have

f (ei � ej) = f (ei � en) = f (�neei) = g (eei) (after (9))


In Case 3, we have

f (ei � ej) = f (en � ej) = f (� (ej � en)) = f (ej � en) (after (8))

= f (�n eej) = g (eej) (after (9))


Thus, f (ei � ej) � 0 holds in all three possible cases. Hence, (10) is proven.Now, our function f : Rn ! R is de�ned by

f (x) =nXu=1

au jxuj �Xs2S

bs jrs nxj ; where x =

0BB@x1x2:::xn

1CCA 2 Rn:

Here, n is a positive integer; the numbers a1; a2; :::; an are n nonnegative reals; theset S is a �nite set; for every s 2 S; the function rs n is an element of (Rn)

� (in otherwords, a linear transformation from Rn to R), and bs is a nonnegative real.Hence, we can apply Theorem 11 to our function f; and we obtain that for our

function f; the Assertions A1 and A2 are equivalent. In other words, our function fsatis�es Assertion A1 if and only if it satis�es Assertion A2:Now, according to (10), our function f satis�es Assertion A2: Thus, this function f

must also satisfy Assertion A1: In other words, f (x) � 0 holds for every x 2 Vn: Hence,f (�nx) � 0 holds for every x 2 Rn�1 (because �nx 2 Vn; since Im�n � Vn). Sincef (�nx) = g (x) according to (9), we have therefore proven that g (x) � 0 holds forevery x 2 Rn�1: Hence, Assertion C1 is proven. Thus, we have showed that C2 =) C1;and thus the proof of Theorem 12b is complete.Since Theorem 12b is equivalent to Theorem 12, this also proves Theorem 12.As if this wasn�t enough, here comes a further restatement of Theorem 12:

Theorem 13. Let n be a nonnegative integer. Let a1; a2; :::; an and a ben+1 nonnegative reals. Let S be a �nite set. For every s 2 S; let rs;1; rs;2;:::; rs;n be n nonnegative reals, and let bs be a nonnegative real. Assumethat the following two conditions hold:

ai + a �Xs2S

bsrs;i for every i 2 f1; 2; :::; ng ;

ai + aj �Xs2S

bs jrs;i � rs;jj for any two distinct integers i and j from f1; 2; :::; ng :

Let y1; y2; :::; yn be n reals. Then,

nXi=1

ai jyij+ a

��nXv=1

yv

��Xs2S

bs

��nXv=1

rs;vyv

�� 0:18

Proof of Theorem 13. For every s 2 S; let rs = (rs;1; rs;2; :::; rs;n) 2 (Rn)� be then-dimensional covector whose i-th coordinate is rs;i for every i 2 f1; 2; :::; ng : De�ne afunction g : Rn ! R by

g (x) =nXu=1

au jxuj+a jx1 + x2 + :::+ xnj�Xs2S


0BB@x1x2:::xn

1CCA 2 Rn:

For every i 2 f1; 2; :::; ng ; we have (ei)u =�1; if u = i;0; if u 6= i

for all u 2 f1; 2; :::; ng ;

so that (ei)1 + (ei)2 + :::+ (ei)n = 1; and for every s 2 S; we have

rsei =

nXu=1

rs;u (ei)u (since rs = (rs;1; rs;2; :::; rs;n))

=nXu=1

rs;u

�1; if u = i;0; if u 6= i

= rs;i;

so that

g (ei) =nXu=1

au j(ei)uj+ a j(ei)1 + (ei)2 + :::+ (ei)nj �Xs2S

bs jrseij

=nXu=1

au

�� 1; if u = i;0; if u 6= i

��+ a j1j �Xs2S

bs jrs;ij

= ai j1j+ a j1j �Xs2S

bs jrs;ij|{z}=rs;i;sincers;i�0

= ai + a�Xs2S

bsrs;i � 0

(since ai + a �Ps2S

bsrs;i by the conditions of Theorem 13).

For any two distinct integers i and j from f1; 2; :::; ng ; we have (ei � ej)u =8<:1; if u = i;�1; if u = j;

0; if u 6= i and u 6= jfor all u 2 f1; 2; :::; ng ; so that (ei � ej)1 + (ei � ej)2 + ::: +

(ei � ej)n = 0; and for every s 2 S; we have

rs (ei � ej) =

nXu=1

rs;u (ei � ej)u (since rs = (rs;1; rs;2; :::; rs;n))

=nXu=1

rs;u

8<:1; if u = i;�1; if u = j;

0; if u 6= i and u 6= j= rs;i � rs;j;

19

and thus

g (ei � ej) =

nXu=1


��+ a��(ei � ej)1 + (ei � ej)2 + :::+ (ei � ej)n

��Xs2S

bs jrs (ei � ej)j

=nXu=1

au

��8<:

1; if u = i;�1; if u = j;

0; if u 6= i and u 6= j

��+ a j0j �Xs2S

bs jrs;i � rs;jj

= (ai j1j+ aj j�1j) + a j0j �Xs2S

bs jrs;i � rs;jj = (ai + aj) + 0�Xs2S

bs jrs;i � rs;jj

= ai + aj �Xs2S

bs jrs;i � rs;jj � 0

(since ai + aj �Ps2S

bs jrs;i � rs;jj by the condition of Theorem 13).

So we have shown that g (ei) � 0 for every integer i 2 f1; 2; :::; ng ; and g (ei � ej) �0 for any two distinct integers i and j from f1; 2; :::; ng : Thus, Assertion B2 of Theorem12 is ful�lled. According to Theorem 12, the assertions B1 and B2 are equivalent, sothat Assertion B1 must be ful�lled as well. Hence, g (x) � 0 for every x 2 Rn: In

particular, if we set x =

0BB@y1y2:::yn

1CCA ; then rsx =nPv=1

rs;vyv (since rs = (rs;1; rs;2; :::; rs;n)),

so that

g (x) =nXu=1

au jyuj+ a jy1 + y2 + :::+ ynj �Xs2S

bs jrsxj

=nXu=1

au jyuj+ a jy1 + y2 + :::+ ynj �Xs2S

bs

��nXv=1

rs;vyv

��=

nXi=1

ai jyij+ a

��nXv=1

yv

��Xs2S

bs

��nXv=1

rs;vyv

�� ;and thus g (x) � 0 yields

nXi=1

ai jyij+ a

��nXv=1

yv

��Xs2S

bs

��nXv=1

rs;vyv

�� 0:Theorem 13 is thus proven.

6. A general condition for Popoviciu-like inequalities

Now, we state a result more general than Theorem 5b:

Theorem 14. Let n be a nonnegative integer. Let a1; a2; :::; an and a ben+1 nonnegative reals. Let S be a �nite set. For every s 2 S; let rs;1; rs;2;

20

:::; rs;n be n nonnegative reals, and let bs be a nonnegative real. Assumethat the following two conditions hold2:

ai + a =Xs2S


ai + aj �Xs2S


Let f be a convex function from an interval I � R to R: Let w1; w2; :::;wn be nonnegative reals. Assume that

nPv=1

wv 6= 0 andnPv=1

rs;vwv 6= 0 for alls 2 S:Let x1; x2; :::; xn be n points from the interval I: Then, the inequality

nXi=1

aiwif (xi)+a

nXv=1

wv

!f

0BB@nPv=1

wvxv

nPv=1

wv

1CCA �Xs2S

bs

nXv=1

rs;vwv

!f

0BB@nPv=1

rs;vwvxv

nPv=1

rs;vwv

1CCAholds.

Remark. Written in a less formal way, this inequality states that

nXi=1

aiwif (xi) + a (w1 + w2 + :::+ wn) f

�w1x1 + w2x2 + :::+ wnxn

w1 + w2 + :::+ wn

��Xs2S

bs (rs;1w1 + rs;2w2 + :::+ rs;nwn) f

�rs;1w1x1 + rs;2w2x2 + :::+ rs;nwnxn

rs;1w1 + rs;2w2 + :::+ rs;nwn

�:

Proof of Theorem 14. Since the elements of the �nite set S are used as labelsonly, we can assume without loss of generality that S = fn+ 2; n+ 3; :::; Ng for someinteger N � n + 1 (we just rename the elements of S into n + 2; n + 3; :::; N; whereN = n+ 1 + jSj ; this is possible because the set S is �nite3). De�ne

ui = aiwi for all i 2 f1; 2; :::; ng ;

un+1 = a

nXv=1

wv

!;

us = �bs

nXv=1

rs;vwv

!for all s 2 fn+ 2; n+ 3; :::; Ng (that is, for all s 2 S).

2The second of these two conditions (ai + aj �Ps2S

bs jrs;i � rs;j j for any two distinct integers i

and j from f1; 2; :::; ng) is identic with the second assumed condition in Theorem 13, but the �rst one(ai + a =

Ps2S

bsrs;i for every i 2 f1; 2; :::; ng) is stronger than the �rst required condition in Theorem

13 (which only said that ai + a �Ps2S

bsrs;i for every i 2 f1; 2; :::; ng).3In particular, N = n+ 1 if S = ?:

21

Also de�ne

zi = xi for all i 2 f1; 2; :::; ng ;

zn+1 =

nPv=1

wvxv

nPv=1

wv

;

zs =

nPv=1

rs;vwvxv

nPv=1

rs;vwv

for all s 2 fn+ 2; n+ 3; :::; Ng (that is, for all s 2 S).

Each of these N reals z1; z2; :::; zN is a weighted mean of the reals x1; x2; :::; xn withnonnegative weights. Since the reals x1; x2; :::; xn lie in the interval I; we can thusconclude that each of the N reals z1; z2; :::; zN lies in the interval I as well. In otherwords, the points z1; z2; :::; zN are N points from I:Now,

nXi=1

aiwif (xi) + a

nXv=1

wv

!f

0BB@nPv=1

wvxv

nPv=1

wv

1CCA�Xs2S

bs

nXv=1

rs;vwv

!f

0BB@nPv=1

rs;vwvxv

nPv=1

rs;vwv

1CCA

=nXi=1

aiwi|{z}=ui

f

0@ xi|{z}=zi

1A+ a

nXv=1

wv

!| {z }

=un+1

f

0BBBBBBB@nPv=1

wvxv

nPv=1

wv| {z }=zn+1

1CCCCCCCA+Xs2S

0BBB@�bs

nXv=1

rs;vwv

!| {z }

=us

1CCCA f

0BBBBBBB@nPv=1

rs;vwvxv

nPv=1

rs;vwv| {z }=zs

1CCCCCCCA=

nXi=1

uif (zi) + un+1f (zn+1) +Xs2S

usf (zs) =nXi=1

uif (zi) + un+1f (zn+1) +NX

s=n+2

usf (zs)

=

NXk=1

ukf (zk) :

Hence, once we are able to show thatNPk=1

ukf (zk) � 0; we will obtain

nXi=1

aiwif (xi) + a

nXv=1

wv

!f

0BB@nPv=1

wvxv

nPv=1

wv

1CCA �Xs2S

bs

nXv=1

rs;vwv

!f

0BB@nPv=1

rs;vwvxv

nPv=1

rs;vwv

1CCA ;

and thus Theorem 14 will be established.

Therefore, in order to prove Theorem 14, it remains to prove the inequalityNPk=1

ukf (zk) �0:

22

We have

NXk=1

uk =nXi=1

ui + un+1 +NX

s=n+2

us =nXi=1

ui + un+1 +Xs2S

us

=

nXi=1

aiwi + a

nXv=1

wv

!+Xs2S

�bs

nXv=1

rs;vwv

!!

=

nXi=1

aiwi + a

nXi=1

wi

!+Xs2S

�bs

nXi=1

rs;iwi

!!

=

nXi=1

aiwi +nXi=1

awi �nXi=1

Xs2S

bsrs;iwi =nXi=1

aiwi + awi �

Xs2S

bsrs;iwi

!

=nXi=1

ai + a�

Xs2S

bsrs;i

!wi

=

nXi=1

0wi

0@ since ai + a =Ps2S

bsrs;i by an assumption of Theorem 14,

and thus ai + a�Ps2S

bsrs;i = 0

1A= 0:

Next, we are going to prove thatNPk=1

uk jzk � tj � 0 holds for every t 2 fz1; z2; :::; zNg :

In fact, let t 2 fz1; z2; :::; zNg be arbitrary. Set yi = wi (xi � t) for every i 2 f1; 2; :::; ng :Then, for all i 2 f1; 2; :::; ng ; we have wi (zi � t) = wi (xi � t) = yi: Furthermore,

zn+1 � t =

nPv=1

wvxv

nPv=1

wv

� t =

nPv=1

wvxv �nPv=1

wv � tnPv=1

wv

=

nPv=1

wv (xv � t)

nPv=1

wv

=

nPv=1

yv

nPv=1

wv

:

Finally, for all s 2 fn+ 2; n+ 3; :::; Ng (that is, for all s 2 S), we have

zs�t =

nPv=1

rs;vwvxv

nPv=1

rs;vwv

�t =

nPv=1

rs;vwvxv �nPv=1

rs;vwv � tnPv=1

rs;vwv

=

nPv=1

rs;vwv (xv � t)

nPv=1

rs;vwv

=

nPv=1

rs;vyv

nPv=1

rs;vwv

:

23

Hence,

NXk=1

uk jzk � tj =nXi=1

ui jzi � tj+ un+1 jzn+1 � tj+NX

s=n+2

us jzs � tj

=

nXi=1

aiwi jzi � tj| {z }=jwi(zi�t)j;since wi�0

+a

nXv=1

wv

!��nPv=1

yv

nPv=1

wv

��+NX

s=n+2

�bs

nXv=1

rs;vwv

!!��nPv=1

rs;vyv

nPv=1

rs;vwv

��=

nXi=1

ai jwi (zi � t)j+ a

nXv=1

wv

! �� nPv=1

yv

��nPv=1

wv

+NX

s=n+2

�bs

nXv=1

rs;vwv

!! �� nPv=1

rs;vyv

��nPv=1

rs;vwv0BB@ here we have pulled thenPv=1

wv andnPv=1

rs;vwv terms out of the modulus

signs, since they are positive (in fact, they are 6= 0 by an assumptionof Theorem 14, and nonnegative because wi and rs;i are all nonnegative)

1CCA=

nXi=1

ai jyij+ a

��nXv=1

yv

��+NX

s=n+2

(�bs)��nXv=1

rs;vyv

�� =nXi=1

ai jyij+ a

��nXv=1

yv

��NX

s=n+2

bs

��nXv=1

rs;vyv

��=

nXi=1

ai jyij+ a

��nXv=1

yv

��Xs2S

bs

��nXv=1

rs;vyv

�� 0by Theorem 13 (in fact, we were allowed to apply Theorem 13 because all the require-ments of Theorem 13 are ful�lled - in particular, we have ai + a �

Ps2S

bsrs;i for every

i 2 f1; 2; :::; ng because we know that ai+ a =Ps2S

bsrs;i for every i 2 f1; 2; :::; ng by an

assumption of Theorem 14).Altogether, we have now shown the following: The points z1; z2; :::; zN are N points

from I: The N reals u1; u2; :::; uN satisfyNPk=1

uk = 0; andNPk=1

uk jzk � tj � 0 holds for

every t 2 fz1; z2; :::; zNg : Hence, according to Theorem 8b, we haveNPk=1

ukf (zk) � 0:

And as we have seen above, onceNPk=1

ukf (zk) � 0 is shown, the proof of Theorem 14 iscomplete. Thus, Theorem 14 is proven.Theorem 14 gives a su¢ cient criterion for the validity of inequalities of the kind

convex combination of f (x1) ; f (x2) ; :::; f (xn)

and f (some weighted mean of x1; x2; :::; xn)

� convex combination of �nitely many f (some other weighted means of x1; x2; :::; xn)�s,

where f is a convex function and x1; x2; :::; xn are n reals in its domain, and where theweights of the weighted mean on the left hand side are positive (those of the weightedmeans on the right hand side may be 0 as well, but still have to be nonnegative). Thiscriterion turns out to be necessary as well:

24

Theorem 14b. Let n be a nonnegative integer. Let w1; w2; :::; wn bepositive reals. Let a1; a2; :::; an and a be n + 1 nonnegative reals. Let Sbe a �nite set. For every s 2 S; let rs;1; rs;2; :::; rs;n be n nonnegative reals,and let bs be a nonnegative real. Let I � R be an interval.Assume that the inequality

nXi=1

aiwif (xi)+a

nXv=1

wv

!f

0BB@nPv=1

wvxv

nPv=1

wv

1CCA �Xs2S

bs

nXv=1

rs;vwv

!f

0BB@nPv=1

rs;vwvxv

nPv=1

rs;vwv

1CCAholds for any convex function f : I ! R and any n points x1; x2; :::; xn inthe interval I: Then,

ai + a =Xs2S


ai + aj �Xs2S


Since we are not going to use this fact, we are not proving it either, but the idea ofthe proof is the following: Assume WLOG that I = [�1; 1] : For every i 2 f1; 2; :::; ng ;you get ai + a �

Ps2S

bsrs;i (by considering the convex function f (x) = x and the

points xk =�1; if k = i;0; if k 6= i

) and ai+ a �Ps2S

bsrs;i (by considering the convex function

f (x) = �x and the same points), so that ai + a =Ps2S

bsrs;i: For any two distinct

integers i and j from f1; 2; :::; ng ; you get ai + aj �Ps2S

bs jrs;i � rs;jj (by considering

the convex function f (x) = jxj and the points xk =

8<:1; if k = i;�1; if k = j;

0; if k 6= i and k 6= j). This

altogether proves Theorem 14b.

7. Proving the Popoviciu inequality

Now we can �nally step to the proof of Theorem 5b:We assume that n � 2; because all cases where n < 2 (that is, n = 1 or n = 0) can

be checked manually (and are uninteresting).

Let ai =�n� 2m� 1

�for every i 2 f1; 2; :::; ng : Let a =

�n� 2m� 2

�: These reals a1; a2;

:::; an and a are all nonnegative (since n � 2 yields n � 2 � 0 and thus�n� 2t

�� 0

for all integers t).Let S = fs � f1; 2; :::; ng j jsj = mg ; that is, we denote by S the set of allm-element

subsets of the set f1; 2; :::; ng : This set S is obviously �nite.For every s 2 S; de�ne n reals rs;1; rs;2; :::; rs;n as follows:

rs;i =

�1; if i 2 s;0; if i =2 s for every i 2 f1; 2; :::; ng :

25

Obviously, these reals rs;1; rs;2; :::; rs;n are all nonnegative. Also, for every s 2 S; setbs = 1; then, bs is a nonnegative real as well.For every i 2 f1; 2; :::; ng ; we haveXs2S

bsrs;i =Xs2S

1rs;i =Xs2S

rs;i =Xs2S

�1; if i 2 s;0; if i =2 s =

Xs�f1;2;:::;ng;

jsj=m

�1; if i 2 s;0; if i =2 s

= (number of m-element subsets s of the set f1; 2; :::; ng that contain i)

=

�n� 1m� 1

�;

so that

ai + a =

�n� 2m� 1

�+

�n� 2m� 2

�=

�n� 1m� 1

�(by the recurrence relation of the binomial coe¢ cients)

=Xs2S

bsrs;i: (11)

For any two distinct integers i and j from f1; 2; :::; ng ; we haveXs2S

bs jrs;i � rs;jj =Xs2S

1 jrs;i � rs;jj =Xs2S

jrs;i � rs;jj

=Xs2S

�� 1; if i 2 s;0; if i =2 s �

�1; if j 2 s;0; if j =2 s

�� =Xs2S

8>><>>:0; if i 2 s and j 2 s;1; if i 2 s and j =2 s;1; if i =2 s and j 2 s;0; if i =2 s and j =2 s

=Xs2S

��1; if i 2 s and j =2 s;

0 otherwise+

�1; if i =2 s and j 2 s;

0 otherwise

�=Xs2S

�1; if i 2 s and j =2 s;

0 otherwise+Xs2S

�1; if i =2 s and j 2 s;

0 otherwise

=X

s�f1;2;:::;ng;jsj=m

�1; if i 2 s and j =2 s;

0 otherwise+

Xs�f1;2;:::;ng;

jsj=m

�1; if i =2 s and j 2 s;

0 otherwise

= (number of m-element subsets s of the set f1; 2; :::; ng that contain i but not j)+ (number of m-element subsets s of the set f1; 2; :::; ng that contain j but not i)

=

�n� 2m� 1

�+

�n� 2m� 1

�= ai + aj;

so thatai + aj =

Xs2S

bs jrs;i � rs;jj : (12)

Also,nXv=1

wv = w1 + w2 + :::+ wn 6= 0 (13)

26

(by an assumption of Theorem 5b).The elements of S are all the m-element subsets of f1; 2; :::; ng : Hence, to every

element s 2 S uniquely correspond m integers i1; i2; :::; im satisfying 1 � i1 < i2 <::: < im � n and s = fi1; i2; :::; img (these m integers i1; i2; :::; im are the m elements ofs in increasing order). And conversely, any m integers i1; i2; :::; im satisfying 1 � i1 <i2 < ::: < im � n can be obtained this way - in fact, they correspond to the m-elementset s = fi1; i2; :::; img 2 S: Given an element s 2 S and the corresponding m integersi1; i2; :::; im; we can write

nXv=1

rs;vwv =

nXv=1

�1; if v 2 s;0; if v =2 s � wv =

Xv2s

wv =X

v2fi1;i2;:::;img

wv = wi1 + wi2 + :::+ wim ;

nXv=1

rs;vwvxv =nXv=1

�1; if v 2 s;0; if v =2 s � wvxv =

Xv2s

wvxv

=X

v2fi1;i2;:::;img

wvxv = wi1xi1 + wi2xi2 + :::+ wimxim :

From this, we can conclude that

nXv=1

rs;vwv 6= 0 for every s 2 S (14)

(becausenPv=1

rs;vwv = wi1+wi2+ :::+wim ; and wi1+wi2+ :::+wim 6= 0 by an assumption

of Theorem 5b), and we can also conclude that

Xs2S

nXv=1

rs;vwv

!f

0BB@nPv=1

rs;vwvxv

nPv=1

rs;vwv

1CCA=

X1�i1<i2<:::<im�n

(wi1 + wi2 + :::+ wim) f


wi1 + wi2 + :::+ wim

�:

(15)

Using the conditions of Theorem 5b and the relations (11), (12), (13) and (14), wesee that all conditions of Theorem 14 are ful�lled. Thus, we can apply Theorem 14,and obtain

nXi=1

aiwif (xi) + a

nXv=1

wv

!f

0BB@nPv=1

wvxv

nPv=1

wv

1CCA �Xs2S

bs

nXv=1

rs;vwv

!f

0BB@nPv=1

rs;vwvxv

nPv=1

rs;vwv

1CCA :

27

This rewrites as

nXi=1

�n� 2m� 1

�wif (xi) +

�n� 2m� 2

� nXv=1

wv

!f

0BB@nPv=1

wvxv

nPv=1

wv

1CCA

�Xs2S

1

nXv=1

rs;vwv

!f

0BB@nPv=1

rs;vwvxv

nPv=1

rs;vwv

1CCA :

In other words,

�n� 2m� 1

� nXi=1

wif (xi) +

�n� 2m� 2

� nXv=1

wv

!f

0BB@nPv=1

wvxv

nPv=1

wv

1CCA

�Xs2S

nXv=1

rs;vwv

!f

0BB@nPv=1

rs;vwvxv

nPv=1

rs;vwv

1CCA :

Using (15) and the obvious relations

nXv=1

wv = w1 + w2 + :::+ wn;

nXv=1

wvxv = w1x1 + w2x2 + :::+ wnxn;

we can rewrite this as�n� 2m� 1

� nXi=1

wif (xi) +

�n� 2m� 2

�(w1 + w2 + :::+ wn) f

�w1x1 + w2x2 + :::+ wnxn

w1 + w2 + :::+ wn

��

X1�i1<i2<:::<im�n

(wi1 + wi2 + :::+ wim) f


wi1 + wi2 + :::+ wim

�:

This proves Theorem 5b.

8. A cyclic inequality

The most general form of the Popoviciu inequality is now proven. But this is notthe end to the applications of Theorem 14. We will now apply it to show a cyclicinequality similar to Popoviciu�s:

Theorem 15a. Let f be a convex function from an interval I � R to R:Let x1; x2; :::; xn be �nitely many points from I:

We extend the indices in x1; x2; :::; xn cyclically modulo n; this meansthat for any integer i =2 f1; 2; :::; ng ; we de�ne a real xi by setting xi = xj;

28

where j is the integer from the set f1; 2; :::; ng such that i � jmodn: (Forinstance, this means that xn+3 = x3:)

Let x =x1 + x2 + :::+ xn

n. Let r be an integer. Then,

2

nXi=1

f (xi) + n (n� 2) f (x) � n

nXs=1

f

�x+

xs � xs+rn

�:

A weighted version of this inequality is:

Theorem 15b. Let f be a convex function from an interval I � R to R:Let x1; x2; :::; xn be �nitely many points from I: Let r be an integer.

Let w1; w2; :::; wn be nonnegative reals. Let x =

nPv=1

wvxv

nPv=1

wv

and w =nPv=1

wv.

Assume that w 6= 0 and that w + (ws � ws+r) 6= 0 for every s 2 S:We extend the indices in x1; x2; :::; xn and in w1; w2; :::; wn cyclicallymodulo n; this means that for any integer i =2 f1; 2; :::; ng ; we de�ne realsxi and wi by setting xi = xj and wi = wj; where j is the integer fromthe set f1; 2; :::; ng such that i � jmodn: (For instance, this means thatxn+3 = x3 and wn+2 = w2:)

Then,

2nXi=1

wif (xi)+(n� 2)wf (x) �nXs=1

(w + (ws � ws+r)) f

0BB@nPv=1

wvxv + (wsxs � ws+rxs+r)

w + (ws � ws+r)

1CCA :

Proof of Theorem 15b. We assume that n � 2; because all cases where n < 2 (thatis, n = 1 or n = 0) can be checked manually (and are uninteresting).Before we continue with the proof, let us introduce a simple notation: For any

assertion A, we denote by [A] the Boolean value of the assertion A (that is, [A] =�1, if A is true;0, if A is false ). Therefore, 0 � [A] � 1 for every assertion A:

Let ai = 2 for every i 2 f1; 2; :::; ng : Let a = n� 2: These reals a1; a2; :::; an and aare all nonnegative (since n � 2 yields n� 2 � 0).Let S = f1; 2; :::; ng : This set S is obviously �nite.For every s 2 S; de�ne n reals rs;1; rs;2; :::; rs;n as follows:

rs;i = 1 + [i = s]� [i � s+ rmodn] for every i 2 f1; 2; :::; ng :

These reals rs;1; rs;2; :::; rs;n are all nonnegative (because

rs;i = 1 + [i = s]| {z }�0

� [i � s+ rmodn]| {z }�1

� 1 + 0� 1 = 0

29

for every i 2 f1; 2; :::; ng). Also, for every s 2 S; set bs = 1; then, bs is a nonnegativereal as well.For every i 2 f1; 2; :::; ng ; we have

nPs=1

[i = s] = 1 (because there exists one and

only one s 2 f1; 2; :::; ng satisfying i = s). Also, for every i 2 f1; 2; :::; ng ; we havenPs=1

[s � i� rmodn] = 1 (because there exists one and only one s 2 f1; 2; :::; ng satisfy-

ing s � i�rmodn). In other words,nPs=1

[i � s+ rmodn] = 1 (because [s � i� rmodn] =

[i � s+ rmodn]).For every i 2 f1; 2; :::; ng ; we haveX

s2Sbsrs;i =

nXs=1

bs|{z}=1

rs;i =nXs=1

rs;i =nXs=1

(1 + [i = s]� [i � s+ rmodn])

=nXs=1

1 +nXs=1

[i = s]�nXs=1

[i � s+ rmodn] = n+ 1� 1 = n = 2 + (n� 2) = ai + a;

so thatai + a =

Xs2S

bsrs;i: (16)

For any two integers i and j from f1; 2; :::; ng ; we havenXs=1

jrs;i � 1j =nXs=1

j(1 + [i = s]� [i � s+ rmodn])� 1j

=nXs=1

j[i = s] + (� [i � s+ rmodn])j �nXs=1

(j[i = s]j+ j� [i � s+ rmodn]j)

(by the triangle inequality)

=nXs=1

([i = s] + [i � s+ rmodn])�because [i = s] and [i � s+ rmodn] are nonnegative, so thatj[i = s]j = [i = s] and j� [i � s+ rmodn]j = [i � s+ rmodn]

�=

nXs=1

[i = s] +

nXs=1

[i � s+ rmodn] = 1 + 1 = 2

and similarlynPs=1

jrs;j � 1j � 2; so that

Xs2S

bs jrs;i � rs;jj =nXs=1

bs|{z}=1

jrs;i � rs;jj =nXs=1

jrs;i � rs;jj =nXs=1

j(rs;i � 1) + (1� rs;j)j

�nXs=1

(jrs;i � 1j+ j1� rs;jj) (by the triangle inequality)

=nXs=1

(jrs;i � 1j+ jrs;j � 1j) =nXs=1

jrs;i � 1j+nXs=1

jrs;j � 1j

� 2 + 2 = ai + aj;

30

and thusai + aj �

Xs2S

bs jrs;i � rs;jj : (17)

For every s 2 S (that is, for every s 2 f1; 2; :::; ng), we havenXv=1

[v � s+ rmodn] � wv =nXv=1

�wv; if v � s+ rmodn;

0 otherwise= ws+r

(because there is one and only one element v 2 f1; 2; :::; ng that satis�es v � s+rmodn;and for this element v; we have wv = ws+r), so that

nXv=1

rs;vwv =nXv=1

(1 + [v = s]� [v � s+ rmodn]) � wv

=nXv=1

wv| {z }=w

+nXv=1

[v = s] � wv| {z }=ws

�nXv=1

[v � s+ rmodn] � wv| {z }=ws+r

= w + ws � ws+r = w + (ws � ws+r) :

Also, for every s 2 S (that is, for every s 2 f1; 2; :::; ng), we havenXv=1

[v � s+ rmodn] � wvxv =nXv=1

�wvxv; if v � s+ rmodn;

0 otherwise= ws+rxs+r

(because there is one and only one element v 2 f1; 2; :::; ng that satis�es v � s+rmodn;and for this element v; we have wv = ws+r and xv = xs+r), and thus

nXv=1

rs;vwvxv =

nXv=1

(1 + [v = s]� [v � s+ rmodn]) � wvxv

=

nXv=1

wvxv +nXv=1

[v = s] � wvxv| {z }=wsxs

�nXv=1

[v � s+ rmodn] � wvxv| {z }=ws+rxs+r

=

nXv=1

wvxv + wsxs � ws+rxs+r =

nXv=1

wvxv + (wsxs � ws+rxs+r) :

Now it is clear thatnPv=1

rs;vwv 6= 0 for all s 2 S (becausenPv=1

rs;vwv = w+(ws � ws+r)

and w+(ws � ws+r) 6= 0). Also,nPv=1

wv 6= 0 (sincenPv=1

wv = w and w 6= 0). Using these

two relations, the conditions of Theorem 15b and the relations (16) and (17), we seethat all conditions of Theorem 14 are ful�lled. Hence, we can apply Theorem 14 andobtain

nXi=1

ai|{z}=2

wif (xi)+ a|{z}=n�2

0BBB@nXv=1

wv| {z }=w

1CCCA f

0BBBBBB@nPv=1

wvxv

nPv=1

wv| {z }=x

1CCCCCCA �Xs2S

bs|{z}=1

nXv=1

rs;vwv

!f

0BB@nPv=1

rs;vwvxv

nPv=1

rs;vwv

1CCA :

31

This immediately simpli�es to

nXi=1

2wif (xi) + (n� 2)wf (x) �Xs2S

1

nXv=1

rs;vwv

!f

0BB@nPv=1

rs;vwvxv

nPv=1

rs;vwv

1CCA :

Recalling that for every s 2 S; we havenPv=1

rs;vwv = w+(ws � ws+r) andnPv=1

rs;vwvxv =

nPv=1

wvxv + (wsxs � ws+rxs+r) ; we can rewrite this as

nXi=1

2wif (xi)+(n� 2)wf (x) �Xs2S

1 (w + (ws � ws+r)) f

0BB@nPv=1


w + (ws � ws+r)

1CCA :

In other words,

2

nXi=1

wif (xi)+(n� 2)wf (x) �nXs=1

(w + (ws � ws+r)) f

0BB@nPv=1


w + (ws � ws+r)

1CCA :

This proves Theorem 15b.Theorem 15a follows from Theorem 15b if we set w1 = w2 = ::: = wn = 1:Theorem 15a generalizes two inequalities that have appeared on the MathLinks

forum. The �rst of these results if we apply Theorem 15a to r = 1; to r = 2; to r = 3;and so on up to r = n� 1; and sum up the n� 1 inequalities obtained:

Theorem 16. Let f be a convex function from an interval I � R to R:Let x1; x2; :::; xn be �nitely many points from I:

Let x =x1 + x2 + :::+ xn

n. Then,

2 (n� 1)nXi=1

f (xi) + n (n� 1) (n� 2) f (x) � nX1�i�n;1�j�n;i6=j

f

�x+

xi � xjn

�:

This inequality occured in [6], post #4 as a result by Vasile Cîrtoaje (Vasc). OurTheorem 15a is therefore a strengthening of this result.The next inequality was proposed by Michael Rozenberg (aka "Arqady") in [7]:

Theorem 17. Let a; b; c; d be four nonnegative reals. Then,

a4 + b4 + c4 + d4 + 4abcd � 2�a2bc+ b2cd+ c2da+ d2ab

�:

32

Proof of Theorem 17. The case when at least one of the reals a; b; c; d equals 0 iseasy (and a limiting case). Hence, we can assume for the rest of this proof that noneof the reals a; b; c; d equals 0: Since a; b; c; d are nonnegative, this means that a; b; c;d are positive.Let A = ln (a4) ; B = ln (b4) ; C = ln (c4) ; D = ln (d4) : Then, expA = a4; expB =

b4; expC = c4; expD = d4:Let I � R be an interval containing the reals A; B; C; D (for instance, I = R). Let

f : I ! R be the function de�ned by f (x) = expx for all x 2 I: Then, it is known thatthis function f is convex. Applying Theorem 15a to n = 4; x1 = A; x2 = B; x3 = C;x4 = D; and r = 3; we obtain

2 (f (A) + f (B) + f (C) + f (D)) + 4 (4� 2) f�A+B + C +D

4

�� 4

�f

�A+B + C +D

4+A�D

4

�+ f

�A+B + C +D

4+B � A

4

�+f

�A+B + C +D

4+C �B

4

�+ f

�A+B + C +D

4+D � C

4

��:

Dividing this by 2 and simplifying, we obtain

f (A) + f (B) + f (C) + f (D) + 4f

�A+B + C +D

4

�� 2

�f

�2A+B + C

4

�+ f

�2B + C +D

4

�+ f

�2C +D + A

4

�+ f

�2D + A+B

4

��:

Since we have

f (A) = expA = a4 and similarly

f (B) = b4; f (C) = c4; and f (D) = d4;

f

�A+B + C +D

4

�= exp

A+B + C +D

4= 4pexpA � expB � expC � expD

=4pa4 � b4 � c4 � d4 = abcd;

f

�2A+B + C

4

�= exp

2A+B + C

4=

4

q(expA)2 � expB � expC

=4

q(a4)2 � b4 � c4 = a2bc and similarly

f

�2B + C +D

4

�= b2cd; f

�2C +D + A

4

�= c2da; and f

�2D + A+B

4

�= d2ab;

this becomes

a4 + b4 + c4 + d4 + 4abcd � 2�a2bc+ b2cd+ c2da+ d2ab

�:

This proves Theorem 17.

References

33

[1] Vasile Cîrtoaje, Two Generalizations of Popoviciu�s Inequality, Crux Mathe-maticorum 5/2001 (volume 31), pp. 313-318.http://journals.cms.math.ca/CRUX/[2] Billzhao et al., Generalized Popoviciu - MathLinks topic #19097.

http://www.mathlinks.ro/Forum/viewtopic.php?t=19097[3] Billzhao et al., Like Popoviciu - MathLinks topic #21786.

http://www.mathlinks.ro/Forum/viewtopic.php?t=21786[4] Darij Grinberg et al., The Karamata Inequality - MathLinks topic #14975.

http://www.mathlinks.ro/Forum/viewtopic.php?t=14975[5] Darij Grinberg et al., Weighted majorization and a result stronger than Fuchs -

MathLinks topic #104714.http://www.mathlinks.ro/Forum/viewtopic.php?t=104714[6] Harazi et al., improvement of Popoviciu�s inequality in a particular case - Math-

Links topic #22364.http://www.mathlinks.ro/Forum/viewtopic.php?t=22364[7] Arqady et al., New, old inequality - MathLinks topic #56040.

http://www.mathlinks.ro/Forum/viewtopic.php?t=56040[8] Albert W. Marshall, Ingram Olkin, Inequalities: Theory of Majorization and Its

Applications, 1979.[9] G. H. Hardy, J. E. Littlewood, G. Pólya, Some simple inequalities satis�ed by

convex functions, Messenger Math. 58, pp. 145-152.

34

Generalizations of Popoviciu™s inequality · This criterion entails Vasile Cîrtoaje™s generalization of the Popoviciu inequal- ity (in its standard and in its weighted forms)

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