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19 PP Chung Minh Bat Dang Thuc

Apr 07, 2018

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  • 8/6/2019 19 PP Chung Minh Bat Dang Thuc

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    PHN 1CC KIN THC CN LU

    1/nh ngha0

    0

    A B A B

    A B A B

    2/Tnh cht+ A>B AB B v B >C CA >+ A>B A+C >B + C+ A>B v C > D A+C > B + D+ A>B v C > 0 A.C > B.C+ A>B v C < 0 A.C < B.C+ 0 < A < B v 0 < C B > 0 A n > B n n+ A > B A n > B n vi n l+ A > B A n > B n vi n chn+ m > n > 0 v A > 1 A m > A n + m > n > 0 v 0 0 BA

    11>

    3/Mt s hng bt ng thc

    + A 2 0 vi A ( du = xy ra khi A = 0 )

    + An 0 vi A ( du = xy ra khi A = 0 )+ 0A vi A (du = xy ra khi A = 0 )

    + - A < A = A

    + A B A B+ + ( du = xy ra khi A.B > 0)

    + BABA ( du = xy ra khi A.B < 0)PHN II

    CC PHNG PHP CHNG MINH BT NG THCPhng php 1 : Dng nh ngha

    Kin thc : chng minh A > B. Ta lp hiu A B > 0Lu dng hng bt ng thc M 2 0 vi MV d 1 x, y, z chng minh rng :

    a) x 2 + y 2 + z 2 xy+ yz + zxb) x 2 + y 2 + z 2 2xy 2xz + 2yz

    c) x 2 + y 2 + z 2 +3 2 (x + y + z)Gii:

    1

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    a) Ta xt hiu : x 2 + y 2 + z 2 - xy yz zx =2

    1.2 .( x 2 + y 2 + z 2 - xy yz

    zx)

    =2

    10)()()( 222 ++ zyzxyx ng vi mi x;y;z R

    V (x-y)2 0 vi

    x ; y Du bng xy ra khi x=y(x-z)2 0 vix ; z Du bng xy ra khi x=z(y-z)2 0 vi z; y Du bng xy ra khi z=yVy x 2 + y 2 + z 2 xy+ yz + zx. Du bng xy ra khi x = y =z

    b)Ta xt hiu: x 2 + y 2 + z 2 - ( 2xy 2xz +2yz ) = x 2 + y 2 + z 2 - 2xy +2xz 2yz

    = ( x y + z) 2 0 ng vi mi x;y;z RVy x 2 + y 2 + z 2 2xy 2xz + 2yz ng vi mi x;y;z RDu bng xy ra khi x+y=zc) Ta xt hiu: x 2 + y 2 + z 2 +3 2( x+ y +z ) = x 2 - 2x + 1 + y 2 -2y +1 + z 2 -

    2z +1= (x-1) 2 + (y-1) 2 +(z-1) 2 0. Du(=)xy ra khi x=y=z=1V d 2: chng minh rng :

    a)222

    22

    +

    + baba; b)

    2222

    33

    ++

    ++ cbacbac) Hy tng qut bi

    tonGii:

    a) Ta xt hiu222

    22

    +

    + baba

    = ( )4

    2

    4

    2 2222 bababa ++

    += ( )abbaba 222

    41 2222 + = ( ) 0

    41 2 ba

    Vy222

    22

    +

    + baba. Du bng xy ra khi a=b

    b)Ta xt hiu

    2222

    33

    ++

    ++ cbacba= ( ) ( ) ( )[ ] 0

    9

    1 222 ++ accbba .Vy

    2222

    33

    ++++ cbacba

    Du bng xy ra khi a = b =c

    c)Tng qut2

    2122

    221 ........

    +++

    +++n

    aaa

    n

    aaa nn

    Tm li cc bc chng minh A B theo nh nghaBc 1: Ta xt hiu H = A - BBc 2:Bin i H=(C+D) 2 hoc H=(C+D) 2 +.+(E+F) 2

    2

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    Bc 3:Kt lun A BV d 1: Chng minh m,n,p,q ta u c : m 2 + n 2 + p 2 + q 2 +1 m(n+p+q+1) Gii:

    014444

    22

    22

    22

    2

    ++

    ++

    ++

    + m

    mqmq

    mpmp

    mnmn

    m

    012222

    2222

    +

    +

    +

    mqmpmnm (lun ng)

    Du bng xy ra khi

    =

    =

    =

    =

    012

    02

    02

    02

    m

    qm

    pm

    nm

    =

    =

    =

    =

    22

    2

    2

    m

    mq

    mp

    mn

    ====

    1

    2

    qpn

    m

    V d 2:Chng minh rng vi mi a, b, c ta lun c : )(444 cbaabccba ++++

    Gii: Ta c : )(444 cbaabccba ++++ , 0,, > cba

    ( ) ( ) ( )

    ( ) ( ) ( )

    ( ) ( ) ( ) ( ) ( ) ( ) 00)2(

    )2()2(

    0222

    222

    0222222

    0

    222222222222

    22222

    2222222222222222222

    222

    222222222222222

    222444

    222444

    +++++

    ++

    ++++++

    +++++

    ++

    ++

    acabacbcbcabaccbba

    abaacba

    abcaccbacbcbbaaccbba

    abcacbbca

    caaccbcbbaba

    abcacbbcacba

    abcacbbcacba

    ng vi mi a, b, c.Phng php 2 : Dng php bin i tng ng

    Kin thc:Ta bin i bt ng thc cn chng minh tng ng vi bt ng thc

    ng hoc bt ng thc c chng minh l ng.Nu A < B C < D , vi C < D l mt bt ng thc hin nhin, hoc bit l ng

    th c bt ng thc A < B .Ch cc hng ng thc sau: ( ) 222 2 BABABA ++=+

    ( ) BCACABCBACBA 2222222

    +++++=++ ( ) 32233 33 BABBAABA +++=+ V d 1: Cho a, b, c, d,e l cc s thc chng minh rng

    a) abb

    a +4

    22

    b) baabba ++++ 122

    c) ( )edcbaedcba +++++++ 22222

    Gii:

    3

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    a) abb

    a +4

    22 abba 44 22 + 044 22 + baa ( ) 02 2 ba

    (BT ny lun ng). Vy abb

    a +4

    22 (du bng xy ra khi 2a=b)

    b) baabba ++++ 122 ) )(21(2 22 baabba ++>++

    0121222222

    +++++ bbaababa 0)1()1()( 222 ++ baba Bt ng thc cui ng.Vy baabba ++++ 122 . Du bng xy ra khi a=b=1c) ( )edcbaedcba +++++++ 22222

    ( ) ( )edcbaedcba +++++++ 44 22222 ( ) ( ) ( ) ( ) 044444444 22222222 +++++++ cacadadacacababa

    ( ) ( ) ( ) ( ) 02222 2222 +++ cadacabaBt ng thc ng vy ta c iu phi chng minh

    V d 2: Chng minh rng: ( )( ) ( )( )4488221010 babababa ++++Gii:( )( ) ( )( )4488221010 babababa ++++

    128448121210221012 bbabaabbabaa ++++++ ( ) ( ) 022822228 + abbababa a2b2(a2-b2)(a6-b6) 0

    a2b2(a2-b2)2(a4+ a2b2+b4) 0Bt ng thccui ng vy ta c iu phi chng minh

    V d 3: cho x.y =1 v x y Chng minhyx

    yx

    + 22

    22

    Gii: yx

    yx

    + 22

    22 v :xy nn x- y

    0

    x

    2

    +y

    2 22 ( x-y)

    x2+y2- 22 x+ 22 y 0 x2+y2+2- 22 x+ 22 y -2 0 x2+y2+( 2 )2- 22 x+ 22 y -2xy 0 v x.y=1 nn 2.x.y=2 (x-y- 2 )2 0 iu ny lun lun ng . Vy ta c iu phi chng

    minhV d 4: Chng minh rng:a/ P(x,y)= 01269 222 ++ yxyyyx Ryx ,

    b/ cbacba ++++ 222 (gi :bnh phng 2 v)c/ Cho ba s thc khc khng x, y, z tha mn:

    ++

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    =(xyz-1)+(x+y+z)-xyz(zyx

    111++ )=x+y+z - ( 0)

    111>++

    zyx(v

    zyx

    111++ < x+y+z

    theo gt) 2 trong 3 s x-1 , y-1 , z-1 m hoc c ba s-1 , y-1, z-1 l dng.

    Nu trng hp sau xy ra th x, y, z >1 x.y.z>1 Mu thun gt x.y.z=1 bt

    buc phi xy ra trng hp trn tc l c ng 1 trong ba s x ,y ,z l s ln hn 1V d 5: Chng minh rng : 21

    ++++

    , )3(cba

    c

    ca

    c

    ++>+

    Cng v theo v cc bt ng thc (1), (2), (3), ta c :

    1>+++++ ca

    c

    cb

    b

    ba

    a(*)

    Ta c : )4(cba

    ca

    ba

    abaa

    +++b>c>0 v 1222 =++ cba . Chng minh rng3 3 3 1

    2

    a b c

    b c a c a b+ +

    + + +Gii:

    Do a,b,c i xng ,gi s ab c

    +

    +

    +

    ba

    c

    ca

    b

    cb

    a cba222

    p dng BT Tr- b-sp ta c

    ++

    ++

    +++

    +

    ++

    ++ ba

    c

    ca

    b

    cb

    acba

    ba

    cc

    ca

    bb

    cb

    aa .

    3...

    222222 =

    2

    3.

    3

    1=

    2

    1

    Vy2

    1333

    ++

    ++

    + bac

    ca

    b

    cb

    aDu bng xy ra khi a=b=c=

    3

    1

    V d 4: Cho a,b,c,d>0 v abcd =1 .Chng minh rng :( ) ( ) ( ) 102222 +++++++++ acddcbcbadcba

    Gii: Ta c abba 222 + cddc 222 +

    Do abcd =1 nn cd =ab

    1(dng

    2

    11+

    xx )

    Ta c 4)1

    (2)(2222 +=+++ab

    abcdabcba (1)

    Mt khc: ( ) ( ) ( )acddcbcba +++++ = (ab+cd)+(ac+bd)+(bc+ad)

    = 222111

    ++

    ++

    ++

    +

    bcbc

    acac

    abab

    Vy ( ) ( ) ( ) 102222 +++++++++ acddcbcbadcba

    Phng php7 Bt ng thc Bernouli Kin thc:

    11

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    a)Dng nguyn thy: Cho a -1, n1 Z th ( ) naa n ++ 11 . Du = xy ra

    khi v ch khi

    =

    =

    1

    0

    n

    a

    b) Dng m rng:- Cho a > -1, 1 th ( ) naa ++ 11 . Du bng xy ra khi v ch khi a = 0.

    - cho 10,1 + baba ab .Gii- Nu 1a hay 1b th BT lun ng- Nu 0 < a,b < 1

    p dng BT Bernouli:( )11 1

    1 1 .b b

    bb aa a b aaa a a a a b

    + = + < + < > + Chng minh tng t:

    ba

    bb a

    +> . Suy ra 1>+ ab ba (pcm).

    V d 2: Cho a,b,c > 0.Chng minh rng5555

    33

    ++++ cbacba . (1)

    Gii

    ( ) 3333

    1

    555

    +++

    +++

    ++

    cba

    c

    cba

    b

    cba

    a

    p dng BT Bernouli:

    ( )cba

    acbcbaacb

    cbaa

    ++++

    ++++=

    ++251213

    55

    (2)

    Chng minh tng t ta uc:

    ( )

    cba

    bac

    cba

    b

    +++

    +

    ++

    251

    35

    (3)

    ( )

    cba

    cba

    cba

    c

    +++

    +

    ++

    251

    35

    (4)

    Cng (2) (3) (4) v theo v ta c

    +++ +++ ++3333

    555

    cbac

    cbab

    cbaa (pcm)

    Ch : ta c bi ton tng qut sau y:Cho .1;0,..., 21 > raaa n Chng minh rng

    r

    nrn

    rr

    n

    aaa

    n

    aaa

    ++++++ ........ 2121 .

    Du = naaa === ....21 .(chng minh tng t bi trn).V d 3:Cho 1,,0 zyx . Chng minh rng

    12

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    ( )( )8

    81222222 ++++ zyxzyx .

    Giit ( )2,,12,2,2 === cbacba zyx .

    ( )( )

    )1(32

    023

    02121

    2

    ++

    aaaa

    aaa

    Chng minh tng t:

    )3(32

    )2(32

    +

    +

    cc

    bb

    Cng (1) (2) (3) v theo v ta c

    ( ) ( )

    )(111

    )(8

    81

    11122

    11129

    pcmcbacba

    cbacba

    cbacba

    csi

    ++++

    ++++

    +++++

    Ch : Bi ton tng qut dng ny Cho n s [ ] 1,,,....,, 21 > cbaxxx nTa lun c:

    ( )( ) ( )[ ]ba

    baxxxxxx

    c

    ccncccccc nn

    +

    +++++++

    4........

    2

    2121

    Phng php 8: S dng tnh cht bc cuKin thc: A>B v B>C th A>C

    V d 1: Cho a, b, c ,d >0 tha mn a> c+d , b>c+dChng minh rng ab >ad+bc

    Gii:

    Tac

    +>+>

    dcb

    dca

    >>>>

    0

    0

    cdb

    dca (a-c)(b-d) > cd

    ab-ad-bc+cd >cd ab> ad+bc (iu phi chng minh)

    V d 2: Cho a,b,c>0 tha mn3

    5222 =++ cba . Chng minh

    abccba

    1111

    0 ta c

    cba

    111+

    abc

    1

    V d 3: Cho 0 < a,b,c,d 1-a-b-c-d Gii:Ta c (1-a).(1-b) = 1-a-b+ab

    13

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    Do a>0 , b>0 nn ab>0 (1-a).(1-b) > 1-a-b (1)Do c 0 ta c (1-a).(1-b) ( 1-c) > 1-a-b-c

    (1-a).(1-b) ( 1-c).(1-d) > (1-a-b-c) (1-d) =1-a-b-c-d+ad+bd+cd (1-a).(1-b) ( 1-c).(1-d) > 1-a-b-c-d (iu phi chng minh)

    V d 4: Cho 0 b

    ath

    cb

    ca

    b

    a

    ++

    >

    b Nu 1+++>

    >

    ..

    222222

    222222222

    V d2 (HS t gii)1/ Cho a,b,c l chiu di ba cnh ca tam gic

    Chng minh rng )(2222

    cabcabcbacabcab ++

  • 8/6/2019 19 PP Chung Minh Bat Dang Thuc

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    x + y = 1

    MN

    O

    MKH

    M

    x

    y

    www.vnmath.com19 Phng php chng minh Bt ng thc

    Th bu = , av = ; )()(. cbccacvu +=Hn na: += abcbccacvuvuvuvu )()(.),cos(... (PCM)V d 2:

    Cho 2n s: niyx ii ,...,2,1,; = tha mn: .111

    =+ ==

    n

    i

    i

    n

    i

    i yx Chng minh rng:

    22

    1

    22 +=n

    i

    ii yx

    Gii:V hnh

    Trong mt phng ta , xt:),( 111 yxM : ),( 21212 yyxxM ++ ;; ),( 11 nnn yyxxM ++++

    Gi thit suy ra n

    M ng thng x + y = 1. Lc :2

    1

    2

    11 yxOM += ,2

    2

    2

    221 yxMM += ,2

    3

    2

    332 yxMM += ,,22

    1 nnnn yxMM +=

    V 1OM 21MM+ 32MM+2

    21 =++ OHOMMM nnn

    + = 2

    2

    1

    22n

    i

    iiyx (PCM)

    Phng php 13: i bin s

    V d1: Cho a,b,c > 0 Chng minh rng2

    3

    ++

    ++

    + bac

    ac

    b

    cb

    a(1)

    Gii: t x=b+c ; y=c+a ;z= a+b ta c a=2

    xzy +; b =

    2

    yxz +; c =

    2zyx +

    ta c (1) z

    zyx

    y

    yxz

    x

    xzy

    222

    ++

    ++

    +

    2

    3

    3111 +++++z

    y

    z

    x

    y

    z

    y

    x

    x

    z

    x

    y ( 6)()() +++++

    z

    y

    y

    z

    z

    x

    x

    z

    y

    x

    x

    y

    18

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    Bt ng thc cui cng ng v ( ;2+y

    x

    x

    y 2+

    z

    x

    x

    z; 2+

    z

    y

    y

    znn ta c

    iu phi chng minh V d2:

    Cho a,b,c > 0 v a+b+c 0 , b > 0 , c > 0 CMR: 81625

    >+

    ++

    ++ ba

    c

    ac

    b

    cb

    a

    2)Tng qut m, n, p, q, a, b >0CMR

    ( ) ( )pnmpnmba

    pc

    ac

    nb

    cb

    ma++++

    ++

    ++

    +2

    2

    1

    Phng php 14: Dng tam thc bc hai

    Kin th: Cho f(x) = ax2 + bx + cnh l 1:

    f(x) > 0,

    0

    0ax

    19

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    www.vnmath.com19 Phng php chng minh Bt ng thc

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    Gii: Bt ng thc cn chng minh tng ng vi ( ) 044.22 322242 >++++ xyxxyyxyx ( ) 0414.)1( 22222 >+++ yxyyxy

    Ta c ( ) ( ) 0161414 2222222 +y vy ( ) 0, >yxf (pcm)

    Phng php 15: Dng quy np ton hcKin thc: chng minh bt ng thc ng vi 0nn > ta thc hin cc bc sau :1 Kim tra bt ng thc ng vi 0nn =2 - Gi s BT ng vi n =k (thay n =k vo BT cn chng minh c

    gi l gi thit quy np )3- Ta chng minh bt ng thc ng vi n = k +1 (thay n = k+1vo BT

    cn chng minh ri bin i dng gi thit quy np)4 kt lun BT ng vi mi 0nn >

    V d1: Chng minh rng : nn

    1

    2

    1

    ....2

    1

    1

    1222

    nNn

    (1)

    Gii: Vi n =2 ta c2

    12

    4

    11

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    V tri (2) 242

    .2

    1111 ++++ +

    +++=

    ++ kkkkkkkk babbaabababa

    042

    1111

    +++

    + ++++ kkkkkk bbaababa

    ( ) ( ) 0. baba kk (3)Ta chng minh (3)

    (+) Gi s a b v gi thit cho a -b a b k

    kk bba ( ) ( ) 0. baba kk

    (+) Gi s a < b v theo gi thit - a

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    n=k ( k ):gi s bt ng thc ng, tc l:))(()( 222

    2

    1

    22

    2

    2

    1

    2

    2211 kkkk bbbaaabababa +++++++++

    n= k+1 . Ta cn chng minh:))(()( 2 1

    22

    21

    21

    22

    21

    2112211 ++++ +++++++++ kkkk bbbaaabababa (1)

    Tht vy: 222122

    2

    2

    1

    22

    2

    2

    1 ).())(()1( baabbbaaaVP kkk +++++++++= +2

    1

    2

    1

    22

    2

    2

    1

    2

    .)( +++++++ kkk babbba ++++++ ++++ 112211112211 22)( kkkkkk bababababababa

    2

    1

    2

    1112 ++++ +++ kkkkkk bababa

    2)( 22211 ++++ kkbababa )( 2211 kkbababa +++ 11 ++ kk ba2

    1

    2

    1.++

    + kk ba2

    112211)( +++++ kk bababa

    Vy (1) c chng minhV d 6: Cho n1 , niRba ii ,...,2,1,, = . Chng minh rng:

    n

    aaa

    n

    aaa nn22

    2

    2

    1221 )(+++

    +++

    Gii:n=1: Bt ng thc lun ngn=k ( k ):gi s bt ng thc ng, tc l:

    k

    aaa

    k

    aaa kk22

    2

    2

    1221 )(+++

    +++

    n= k+1 . Ta cn chng minh:1

    )1

    (2

    1

    2

    2

    2

    12121

    ++++

    +

    +++ ++k

    aaa

    k

    aaa kk (1)

    t:k

    aaaa k 132 +

    +++=

    )2(1

    1)1(

    1

    222

    1akaaka

    kVP

    +++=

    +++++

    ++++

    + ++

    k

    aaakak

    k

    aaaka

    k

    kk

    2

    1

    2

    3

    2

    22

    1

    2

    1

    2

    3

    2

    222

    12.

    )1(

    1

    1

    2

    1

    2

    2

    2

    1

    ++++

    = +k

    aaa k

    Vy (1) c chng minh

    V d 7: Chng minh rng: 2,,)1( 1 +> nnnn nn

    Gii: n=2

    =+

    =

    3)1(

    4

    1n

    n

    n

    n 1)1( +> nn nn

    n=k 2 : gi s bt ng thc ng, tc l: 1)1( +> kk kkn= k+1:Ta c : 111 )1()1()1( ++ +++ kkkk kkkk

    212222)1(])1[()1()1( ++=++= kkkk kk

    )2()2(212 kkkk k ++> (v kkkkk 212)1( 222 +>++=+ )

    kkkk )2( + kk kk )2()1( 1 +>+ + Bt ng thc ng vi n= k+1

    Vy 2,,)1( 1 +> nnnn nn

    23

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    V d 8: Chng minh rng: Rxnxnnx ,,sinsinGii: n=1: Bt ng thc lun ngn=k :gi s bt ng thc ng, tc l: xkkx sinsin n= k+1 . Ta cn chng minh: xkxk sin)1()1sin( ++

    Ta c:

    ++

    Rxxx

    Rbababa

    ,1c o s,s i n

    ,,

    Nn: xkxxkxxk sincoscossin)1sin( +=+xkxxkx sin.coscos.sin + xkx sin..sin + xxk sin..sin + xk sin)1( +=

    Bt ng thc ng vi n= k+1. Vy: Rxnxnnx ,,sinsin + Phng php 16: Chng minh phn chngKin thc:1) Gi s phi chng minh bt ng thc no ng , ta hy gi s bt

    ng thc sai v kt hp vi cc gi thit suy ra iu v l , iu v l c thl iu tri vi gi thit , c th l iu tri ngc nhau .T suy ra bt ng thccn chng minh l ng

    2) Gi s ta phi chng minh lun p qMun chng minh qp (vi p : gi thit ng, q : kt lun ng) php

    chng minh c thc hin nh sau:Gi s khng c q ( hoc q sai) suy ra iu v l hoc p sai. Vy phi c

    q (hay q ng)Nh vy ph nh lun ta ghp tt c gi thit ca lun vi ph nh

    kt lun ca n .

    Ta thng dng 5 hnh thc chng minh phn chng sau :A - Dng mnh phn o : P QB Ph nh ri suy tri gi thitC Ph nh ri suy tri vi iu ngD Ph nh ri suy ra 2 iu tri ngc nhauE Ph nh ri suy ra kt lun :

    V d 1: Cho ba s a,b,c tha mn a +b+c > 0 , ab+bc+ac > 0 , abc > 0Chng minh rng a > 0 , b > 0 , c > 0

    Gii:Gi s a 0 th t abc > 0 a 0 do a < 0. M abc > 0 v a < 0 cb 0 a(b+c) > -bc > 0V a < 0 m a(b +c) > 0 b + c < 0

    a < 0 v b +c < 0 a + b +c < 0 tri gi thit a+b+c > 0Vy a > 0 tng t ta c b > 0 , c > 0V d 2:Cho 4 s a , b , c ,d tha mn iu kin

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    ac 2.(b+d) .Chng minh rng c t nht mt trong cc bt ng thc sau lsai: ba 42 < , dc 42 0Trong ba s x-1 , y-1 , z-1 ch c mt s dngTht vy nu c ba s dng th x,y,z > 1 xyz > 1 (tri gi thit)Cn nu 2 trong 3 s dng th (x-1).(y-1).(z-1) < 0 (v l)Vy c mt v ch mt trong ba s x , y,z ln hn 1V d 4: Cho 0,, >cba v a.b.c=1. Chng minh rng: 3++ cba (Bt ng

    thc Cauchy 3 s)

    Gii: Gi s ngc l i:3

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    acb < 22)( bac > 0))(( >+++ cbacba (2)bac < 22)( cba > 0))(( >+++ cbacba (3)

    Nhn (1), (2) v (3) v vi v ta c:0)])()([(

    2 >++++ cbacbacba V l. Vy bi ton c chng minh

    Phng php 17 : S dng bin i lng gic1. Nu Rx th t x = Rcos , [ ] ,0 ; hoc x = Rsin

    2,

    2,

    2. Nu Rx th t x =cos

    R [ )

    2

    3,,0

    c

    3.Nu ( ) ( ) )0(,222 >=+ Rbyax th t )2(,

    s i n

    c o s

    =

    +=

    +=

    Rby

    Rax

    4. Nu 0,222

    >=

    +

    baRb

    y

    a

    x th t )2(,s i n

    c o s

    =

    +=

    +=

    b Ry

    a Rx

    5. Nu trong bi ton xut hin biu thc : ( ) ( )0,,22 >+ babax

    Th t:

    =

    2,

    2,

    tg

    a

    bx

    V d 1: Cmr : ( )( )( ) [ ]1,1,,211311 2222 ++ baabababbaGii : 1,1 ba

    t :

    =

    =

    c o s

    c o

    b

    a [ ]( ) ,0,

    Khi :

    ( ) ( )( )( )

    [ ]

    2 2 2 21 1 3 1 1

    cos .sin cos .sin 3 cos .cos sin .sin

    sin( ) 3.cos( ) 2cos( ) 2,2 ( )6

    a b b a ab b a

    dpcm

    + +

    = + +

    = + + + = +

    V d 2 : Cho 1, ba .Chng minh rng : ababba + 11Gii :

    26

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    t :

    =

    =

    2,0,

    cos

    1co s

    1

    2

    2

    b

    a

    22 2

    2 2 2 2 2 2

    2 2 2 2 2 2

    1 1 ( .cos .co1 1

    cos cos cos cos cos .cos1 (sin 2 sin 2 ) sin( )cos ( ) 1

    2 cos .cos cos .cos cos .cos

    tg tg tg tg a b b a tg tg

    inab

    + + = + = + =

    + + = = =

    V d 3: Cho 0ab .Chng minh rng : 2224

    )4(222

    22

    22

    +

    ba

    baa

    Gii :t:

    =

    22,

    2,2

    btga

    2 2 2 22

    2 2 2

    ( 4 ) ( 2)4( 1).cos

    4 1

    2sin 2 2(1 cos 2 ) 2(sin 2 cos 2 ) 22 2 sin(2 ) 2 2 2 2, 2 2 2

    2

    a a b tg tg tg

    a b tg

    = =

    + +

    = + = =

    Phng php 18: S dng khai trin nh thc Newton.Kin thc:Cng thc nh thc Newton

    ( ) RbaNnbaCban

    k

    kknkn

    n + =

    ,,, *

    0

    .

    Trong h s )0(!)!(

    !nk

    kkn

    nC

    k

    n

    = .

    Mt s tnh cht t bit ca khai trin nh thc Newton:+ Trong khai trin (a + b)n c n + 1 s hng.+ S m ca a gim dn t n n 0, trong khi s m ca b tng t 0 n n.

    Trong mi s hng ca khai trtin nh thc Newton c tng s m ca a v b bngn.

    +Cc h s cch u hai u th bng nhau knn

    kn CC

    = .+ S hng th k + 1 l )0(. nkbaC kknkn

    V d 1:

    Chng minh rng ( )*,0,11 Nnanaa n ++ (bt ng thc bernoulli)

    Gii

    Ta c: ( ) naaCCaCa nnn

    k

    kk

    n

    n +=+=+ =

    11 10

    0

    (pcm)

    V d 2:Chng minh rng:

    27

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    a) *,0,,22

    Nnbababa

    nnn

    +

    +

    b) *,0,,,33

    Nncbacbacba

    nnnn

    ++

    ++

    Gii

    Theo cng thc khai trin nh thc Newton ta c:( )( )

    ( )

    ( )( )( )

    n

    baba

    baCCCCba

    abCbaCbaCbaCba

    bababababa

    niba

    abCabbaCabbaCbaCba

    aCabCabCbCba

    bCbaCbaCaCba

    nnn

    nnnnn

    nnnn

    nn

    nnnn

    nnnn

    nnn

    nnn

    n

    iniiinnniiinin

    nnnn

    nnnn

    nnn

    nnn

    n

    nnn

    nnn

    nn

    nn

    n

    nnn

    nnn

    nn

    nn

    n

    +

    +

    +=+++++=

    +++++++++

    +

    =

    ++++++++=+

    ++++=+

    ++++=+

    2

    )(2)....)((

    )()(....)()(2

    0

    :1,...,2,1,0,

    )()..(....)()(2

    .....

    .....

    110

    110

    1111110

    11110

    11110

    b) t 03

    ++= cbad

    Theo cu (a) ta c:

    n

    nnnn

    nnnnnnnnn

    nn

    nn

    nn

    nnnn

    cbad

    cba

    dcbaddcba

    ddcba

    dcba

    dcba

    dcba

    ++=

    ++

    +++++

    +++

    ++

    +

    =

    ++

    +

    +++

    33

    34

    )4

    (2

    22

    4

    22

    22

    4

    Phng php 19: S dng tch phnHm s: [ ] Rbagf ,:, lin tc, lc :

    * Nu [ ]baxxf ,,0)( th b

    a

    dxxf 0)(

    * Nu [ ]baxxgxf ,),()( th b

    a

    b

    a

    dxxgdxxf )()(

    * Nu [ ]baxxgxf ,),()( v [ ] )()(:, 000 xgxfbax > th

    b

    a

    b

    a

    dxxgdxxf )()( .

    * b

    a

    b

    a

    dxxfdxxf )()( .

    28

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    * Nu [ ]baxMxfm ,,)( th b

    a

    Mdxxfab

    m )(1

    (m, M l hng

    s)V d 1: Cho A, B, C l ba gc ca tam gic.

    Chng minh rng: 3222

    ++C

    tgB

    tgA

    tg

    Gii:

    t ),0(,2

    )( = xx

    tgxf

    ),0(,0)2

    1(22

    1)(

    )2

    1(2

    1)(

    2''

    2'

    >+=

    +=

    xx

    tgx

    tgxf

    xtgxf

    p dng bt ng thc Jensen cho:

    3222

    63

    222

    63

    222

    33

    )()()(

    ++

    ++

    ++++

    ++

    ++

    Ctg

    Btg

    Atg

    tgC

    tgB

    tgA

    tg

    CBAtg

    Ctg

    Btg

    Atg

    CBAf

    CfBfAf

    V d 2: Chng minh:6cos2510

    2

    0

    2

    xdx

    Gii

    Trn on

    2,0 ta c:

    ( )

    2 2 2

    22

    2 2

    2 20 0

    0 cos 1 0 2cos 2 2 2cos 0

    1 1 13 5 2cos 5

    5 5 2cos 3

    1 10 0

    5 2 5 2cos 3 2 10 5 2cos 6

    x x x

    xx

    dx dxpcm

    x x

    PHN III : CC BI TP NNG CAO

    *Dng nh ngha1) Cho abc = 1 v 363 >a . . Chng minh rng +

    3

    2ab2+c2> ab+bc+ac

    Gii: Ta xt hiu: +3

    2ab2+c2- ab- bc ac = +

    4

    2a+

    12

    2ab2+c2- ab- bc ac

    = ( +4

    2ab2+c2- ab ac+ 2bc) +

    12

    2a3bc =(

    2

    a-b- c)2 +

    a

    abca

    12

    363

    29

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    =(2

    a-b- c)2 +

    a

    abca

    12

    363 >0 (v abc=1 v a3 > 36 nn a >0 )

    Vy : +3

    2ab2+c2> ab+bc+ac iu phi chng minh

    2) Chng minh rng

    a) )1.(212244

    +++++ zxxyxzyxb) vi mi s thc a , b, c ta c036245

    22 >+++ baabbac) 024222 22 +++ baabba

    Gii:a) Xt hiu: xxzxyxzyx 22221 222244 ++++ = ( ) ( ) ( ) 22222 1++ xzxyx

    = HH 0 ta c iu phi chng minhb) V tri c th vit H = ( ) ( ) 1112 22 +++ bba H > 0 ta c pcm

    c) v tri c th vit H = ( ) ( )22

    11 ++ bba H 0 ta c iu phi chngminh

    * Dng bin i tng ng1) Cho x > y v xy =1 .Chng minh rng

    ( )( )

    82

    222

    +

    yx

    yx

    Gii: Ta c ( ) ( ) 22 2222 +=+=+ yxxyyxyx (v xy = 1)

    ( ) ( ) ( ) 4.4 24222 ++=+ yxyxyxDo BT cn chng minh tng ng vi ( ) ( ) ( ) 224 .844 yxyxyx ++

    ( ) ( ) 044 24 + yxyx ( )[ ] 02 22 yxBT cui ng nn ta c iu phi chng minh

    2) Cho xy 1 .Chng minh rng

    xyyx +

    ++

    + 12

    1

    1

    1

    122

    Gii:

    Ta c xyyx ++++ 12

    1

    1

    1

    122 01

    1

    1

    1

    1

    1

    1

    1222

    +++

    ++ xyyyx

    ( ) ( ) ( ) ( ) 01.11.1 22

    2

    2

    ++

    +

    ++

    xyy

    yxy

    xyx

    xxy ( ) ( ) ( ) ( ) 01.1)(

    1.1

    )(22

    ++

    +

    ++

    xyy

    yxy

    xyx

    xyx

    ( ) ( )

    ( ) ( ) ( ) 01.1.11

    22

    2

    +++

    xyyx

    xyxyBT cui ny ng do xy > 1 .Vy ta c pcm

    * Dng bt ng thc ph

    30

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    1) Cho a , b, c l cc s thc v a + b +c =1 Chng minh rng3

    1222 ++ cba

    Gii: p dng BT BunhiaCpski cho 3 s (1,1,1) v (a,b,c)Ta c ( ) ( ) ( )2222 .111.1.1.1 cbacba ++++++ ( ) ( )2222 .3 cbacba ++++

    3

    1222 ++ cba (v a+b+c =1 ) (pcm)

    2) Cho a,b,c l cc s dng . Chng minh rng ( ) 9111

    .

    ++++

    cbacba

    (1)

    Gii: (1) 9111 ++++++++a

    c

    a

    c

    c

    b

    a

    b

    c

    a

    b

    a

    93

    ++

    ++

    ++

    b

    c

    c

    b

    a

    c

    c

    a

    a

    b

    b

    a

    p dng BT ph 2+ xy

    y

    x

    Vi x,y > 0. Ta c BT cui cng lun ng

    Vy ( ) 9111

    .

    ++++

    cbacba (pcm)

    * Dng phng php bc cu1) Cho 0 < a, b,c

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    d a d a d a c

    a b c d d a b a b c d

    + + + +< 0 ,p dng BT Csi ta c

    x+ y + z 33 xyz 31 1

    3 27 xyz xyz

    p dng bt ng thc Csi cho x+y ; y+z ; x+z ta c

    ( ) ( ) ( ) ( ) ( ) ( )3. . 3 . .x y y z z x x y y z x z+ + + + + + ( ) ( ) ( )32 3 . .x y y z z x + + +

    Du bng xy ra khi x=y=z=1

    3

    Vy S 8 1 8

    .27 27 729

    = . Vy S c gi tr ln nht l8

    729khi x=y=z=

    1

    3 V d 3: Cho xy+yz+zx = 1. Tm gi tr nh nht ca 4 4 4 x y z+ + Gii: p dng BT Bunhiacpski cho 6 s (x,y,z) ;(x,y,z)

    Ta c ( ) ( ) 22 2 2 2 xy yz zx x y z+ + + + ( ) 22 2 21 x y z + + (1)

    p dng BT Bunhiacpski cho (2 2 2

    , ,x y z ) v (1,1,1)Ta c 2 2 2 2 2 2 2 4 4 4 2 2 2 2 4 4 4( ) (1 1 1 )( ) ( ) 3( )x y z x y z x y z x y z+ + + + + + + + + +

    T (1) v (2) 4 4 41 3( ) x y z + + 4 4 41

    3 x y z + +

    Vy 4 4 4 x y z+ + c gi tr nh nht l1

    3khi x=y=z= 3

    3

    V d 4: Trong tam gic vung c cng cnh huyn , tam gic vung no cdin tch ln nht Gii: Gi cnh huyn ca tam gic l 2a

    ng cao thuc cnh huyn l hHnh chiu cc cnh gc vung ln cnh huyn l x,y

    Ta c S = ( ) 21

    . . . . .2

    x y h a h a h a xy+ = = =

    V a khng i m x+y = 2a. Vy S ln nht khi x.y ln nht x y =Vy trong cc tam gic c cng cnh huyn th tam gic vung cn c din tch

    ln nht2/ Dng Bt ng thc gii phng trnh v h phng trnh

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    V d 1:Gii phng trnh: 2 2 24 3 6 19 5 10 14 4 2 x x x x x x+ + + + + = Gii : Ta c 23 6 19x x+ + 23.( 2 1) 16x x= + + + 23.( 1) 16 16x= + +

    ( )225 10 14 5. 1 9 9 x x x+ + = + +

    Vy 2 24. 3 6 19 5 10 14 2 3 5 x x x x+ + + + + + =

    Du ( = ) xy ra khi x+1 = 0

    x = -1Vy 2 2 24 3 6 19 5 10 14 4 2 x x x x x x+ + + + + = khi x = -1Vy phng trnh c nghim duy nht x = -1

    V d 2: Gii phng trnh 2 22 4 4 3 x x y y+ = + + Gii : p dng BT BunhiaCpski ta c :

    ( )2 2 2 2 22 1 1 . 2 2. 2 2 x x x x+ + + = Du (=) xy ra khi x = 1

    Mt khc ( )224 4 3 2 1 2 2 y y y+ + = + + Du (=) xy ra khi y = -

    1

    2

    Vy2 2

    2 4 4 3 2 x x y y+ = + + = khi x =1 v y =-

    1

    2

    Vy nghim ca phng trnh l1

    1

    2

    x

    y

    =

    =

    V d 3:Gii h phng trnh sau: 4 4 41 x y z

    x y z xyz

    + + = + + =

    Gii: p dng BT Csi ta c4 4 4 4 4 4

    4 4 4 2 2 2 2 2 2

    2 2 2 2 2 2 2 2 2 2 2 2

    x2 2 2

    2 2 2

    x y y z z x y z x y y z z x

    x y y z z y z z x z y x

    + + ++ + = + + + +

    + + + + +

    2 2 2 .( ) y xz z xy x yz xyz x y z + + + +

    V x+y+z = 1) Nn 4 4 4 x y z xyz+ + Du (=) xy ra khi x = y = z =1

    3

    Vy 4 4 41 x y z

    x y z xyz

    + + = + + =

    c nghim x = y = z =1

    3

    V d 4: Gii h phng trnh sau2

    2

    4 8

    2

    xy y

    xy x

    =

    = +

    (1)

    (2)

    T phng trnh (1) 28 0y hay 8y

    T phng trnh (2) 2 2 . 2 2 x x y x + =

    2 2 22 2 2 0 ( 2) 0 2 2 x x x x x + = =

    Nu x = 2 th y = 2 2Nu x = - 2 th y = -2 2

    34

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    Vy h phng trnh c nghim2

    2

    x

    y

    =

    = v

    2 2

    2 2

    x

    y

    =

    = 3/ Dng BT gii phng trnh nghim nguyn

    V d 1: Tm cc s nguyn x,y,z tho mn 2 2 2 3 2 3 x y z xy y z+ + + +

    Gii: V x,y,z l cc s nguyn nn 2 2 2 3 2 3 x y z xy y z+ + + +

    ( )2 2

    2 2 2 2 233 2 3 0 3 3 2 1 04 4

    y y x y z xy y z x xy y z z

    + + + + + + + +

    ( )2 2

    23 1 1 0

    2 2

    y yx z

    + +

    (*)

    M ( )2 2

    23 1 1 0

    2 2

    y yx z

    + +

    ,x y R

    ( )2 2

    23 1 1 02 2y yx z + + =

    02 1

    1 0 22

    11 0

    yx

    xy

    y

    zz

    ==

    = = = =

    Cc s x,y,z phi tm l

    1

    21

    x

    yz

    =

    = =

    V d 2: Tm nghim nguyn dng ca phng trnh1 1 1

    2 x y z

    + + =

    Gii: Khng mt tnh tng qut ta gi s x y z

    Ta c1 1 1 3

    2 2 3z x y z z

    = + +

    M z nguyn dng vy z = 1. Thay z = 1 vo phng trnh ta c1 1

    1

    x y

    + =

    Theo gi s x y nn 1 =1 1

    x y+

    1

    y 2y m y nguyn dng

    Nn y = 1 hoc y = 2Vi y = 1 khng thch hpVi y = 2 ta c x = 2Vy (2 ,2,1) l mt nghim ca phng trnh

    35

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    www.vnmath.com19 Phng php chng minh Bt ng thc

    Hon v cc s trn ta c cc nghim ca phng trnh l (2,2,1);(2,1,2);(1,2,2)

    V d 3:Tm cc cp s nguyn tho mn phng trnh x x y+ = (*)Gii:

    (*) Vi x < 0 , y < 0 th phng trnh khng c ngha

    (*) Vi x > 0 , y > 0Ta c x x y+ = 2 x x y + = 2 0 x y x = >

    t x k= (k nguyn dng v x nguyn dng )Ta c 2.( 1)k k y+ =

    Nhng ( ) ( )22 1 1k k k k < + < + 1k y k < < +

    M gia k v k+1 l hai s nguyn dng lin tip khng tn ti mt snguyn dng no c

    Nn khng c cp s nguyn dng no tho mn phng trnh .

    Vy phng trnh c nghim duy nht l :0

    0

    x

    y

    = =

    Bi tp ngh :

    Bi 1:Chng minh rng vi mi a,b,c > 0 :cbaab

    c

    ac

    b

    bc

    a 111++++

    HD : Chuyn v quy ng mu a v tng bnh phng cc ng thc.

    Bi 2:Chng minh bt ng thc : *)(1)1(

    1..

    4.3

    1

    3.2

    1

    2.1

    1Nn

    nn 0 v a + b + c 1. Cmr : 641

    1

    1

    1

    1

    1

    +

    +

    + cba

    HD : p dng bt ng thc Csi cho

    +

    +

    +

    cba

    11,

    11,

    11

    Bi 4 : Cho 0,0 cbca . Cmr : abcbccac + )()(

    HD : p dng bt ng thc Csi chob

    cb

    a

    c

    a

    ca

    b

    c , , ri cng hai v

    theo v.

    Bi 5: Cho a, b >1. Tm GTNN ca S =11

    22

    +

    ab

    b

    a

    HD : p dng bt ng thc Csi cho 1,1

    22

    abba v xt trng hp du= xy ra .

    Bi 9 : Tm GTLN v GTNN ca y = 2242

    )21(

    1283

    x

    xx

    +++

    HD: t x=

    2,

    2,

    2

    1 tg

    Bi 10: Cho 36x .916 22 =+ y Cmr :4

    2552

    4

    15+ xy

    36

  • 8/6/2019 19 PP Chung Minh Bat Dang Thuc

    37/37

    www.vnmath.com19 Phng php chng minh Bt ng thc

    HD: t :

    =

    =

    sin4

    3

    cos2

    1

    y

    x

    Bi 11: Cmr : [ ]1,1),121(2

    1122 ++ xx

    xx

    HD : t x =

    4,

    4,2sin

    Bi 12: Cho 1,0, cba . Chng minh rng: accbbacba 222222 1 +++++Bi 13: Cho ABC c a, b, c l di cc cnh. Chng minh rng:

    0)()()(222 ++ acaccbcbbaba

    Bi 14: Cho 0,,1, bann . Chng minh rngnnn baba

    ++

    22

    Bi 15: nn 2, . Chng minh rng: 31

    12