tonghop bdt cua mathscope

8/6/2019 tonghop bdt cua mathscope

1/116


2/116


3/116

(b c)2bc

+(c a)2

ca+

(a b)2ab

8.(a b)2 + (b c)2 + (c a)2

(a + b + c)2

(b c)2 a(a + b + c)2 8abc 0Khng mt tnh tng qut gi s a b cTa c:

Sb + Sa = (a + b)(a + b + c)2 16abc 4c(a + b)2 16abc 0

Sb + Sc = (b + c)(a + b + c)2 16abc 4a(b + c)2 16abc 0

2Sb Sb + Sc 0 Sb 0

Nn theo nh l S.O.S ta c iu phi chng minh.ng thc xy ra khi v ch khi a = b = c. 2

230

Mc lc

Li ni u 4

Cc thnh vin tham gia bin son 5

1 Cc bt ng thc kinh in 6

1.1 Bt ng thc gia trung bnh cng v trung bnh nhn (AM-GM). . . . . . . . . 6

1.2 Bt ng thc gia trung bnh cng v trung bnh iu ho (AM-HM). . . . . . . 61.3 Bt ng thc Cauchy - Schwarz. . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4 Bt ng thc Holder. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.5 Bt ng thc Chebyshev. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.6 Bt ng thc Minkowski. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.7 Bt ng thc Schur. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.8 Bt ng thc Vornicu - Schur. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.9 Bt ng thc Bernoulli. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.10 Ba tiu chun SOS thng gp. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Mt s nh gi quen thuc 9

3 Tuyn tp bt ng thc 10

3.1 Bi 1.1 n bi 1.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3.2 Bi 2.1 n bi 2.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.3 Bi 3.1 n bi 3.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.4 Bi 4.1 n bi 4.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

3.5 Bi 5.1 n bi 5.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

3.6 Bi 6.1 n bi 6.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

3.7 Bi 7.1 n bi 7.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1443.8 Bi 8.1 n bi 8.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

3.9 Bi 9.1 n bi 9.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

3.10 Bi 10.1 n bi 10.40 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

3


4/116

Li ni u

Bin vn mi nhp nh vi nhng con sng dt vo b, thuyn vn mi lnh nh theo tng consng i vo i dng, v trong t lin cuc sng vn c nhiu bt cp cn ang xy ra, . . . , ttc nhng iu u l cc bt ng thc trong phm tr c th ca tng lnh vc. Trong tonhc cng vy ni n bt ng thc l chng ta ni n mt lp bi ton kh m n cha bntrong c nhiu li gii p l k lm say m bit bao nhiu ngi.

Trong thi i cng ngh thng tin vi vic kt ni internet bn c th giao lu hc hi c rt

nhiu v cc phng php lm bi bt ng thc, hoc hc hi vi nhiu cun sch v bt ngthc ang by bn trn th trng nhng c mt cun sch bt ng thc hay vi s hi ttinh hoa kin thc ca nhiu ngi th iu chnh l im mnh ca cun sch bt ng thcm cc bn ang cm trn tay.

"Tuyn Tp Bt ng Thc" vi khong bn trm bi ton bt ng thc chn lc c gi tit cc bn tr, cc thy c gio yu ton trn mi min ca t quc, bao gm cc bi tonbt ng thc mi sng to, cc bi ton bt ng thc kh, cc bi ton bt ng thc hay vth v m cc bn tr mun chia s vi mi ngi. iu to nn s hp dn, tnh cp nht vthi i ca cun sch ny.

Bn c hy nhm nhi vi nhng li gii hay, nhng tng c o, nhng sng kin l k trongcch gii tng bi ton t rt kinh nghim hc tp cho mnh, gip cho bn thm yu, thm

tin vo vic gii nhiu bi ton bt ng thc.Vi tinh thn lm vic nghim tc, ham hc hi nhm bin tp xin c gi li cm n susc ti tt c cc bn tham gia gi bi v gii bi, ng thi cng xin by t s cmn v knh trng ti thy gio Chu Ngc Hng - THPT Ninh Hi - Ninh Thun nhittnh c vn k thut latex. Nhm bin tp cng xin gi li cm n ti ban qun tr din nhttp://forum.mathscope.org/index.php c v, ng vin anh em trong qu trnh lm vic ngy hm nay chng ta c mt cun sch hay, c gi tr cao v kin thc chuyn mn m li honton min ph v ti chnh.

"TUYN TP BT NG THC" chnh thc c pht hnh trn cng ng mng nhngngi yu ton, t thi mt lung gi mi em li nhiu iu mi l cho hc sinh, l tiliu tham kho hu ch cho gio vin trong vic ging dy v hc tp bt ng thc.

Do thi gian gp rt v trnh c hn, d rt c gng song nhng sai st l kh trnh khi rtmong nhn c s thng cm, chia s, gp ca cc bn nhm bin tp hon thin cun schtt hn. Mi kin ng gp xin gi v a ch hoangquan9@gmail .

Thay mt nhm bin son, ti xin chn thnh cm n!

H Ni, ngy 10 thng 8 nm 2011

i din nhm bin sonCh bin

Hong Minh Qun-Batigoal

4

Li gii.

S dng bt ng thc AM-GM ta c:

16. 16

32a(a + b)(a + b + c)

3(a + b + c + d)3

= 16. 16

2a

a + b.

3(a + b)

2(a + b + c).

3(a + b)

2(a + b + c).

4(a + b + c)

3(a + b + c + d).

4(a + b + c)

3(a + b + c + d).

4(a + b + c)

3(a + b + c + d)

2aa + b

+ 2.3(a + b)

2(a + b + c)+ 3.

4(a + b + c)

3(a + b + c + d)+ 10

= 2aa + b

+ 3(a + b)a + b + c

+ 4(a + b + c)a + b + c + d

+ 10. (1)

Mt khc, cng theo bt ng thc AM-GM th:

4. 4

24bcd

(a + b)(a + b + c)(a + b + c + d)= 4. 4

2b

a + b.

3c

a + b + c,

4d

a + b + c + d

2ba + b

+3c

a + b + c+

4d

a + b + c + d+ 1. (2)

Cng v theo v (1) v (2) suy ra iu phi chng minh.ng thc xy ra khi v ch khi a = b = c = d. 2

10.39 Cho x,y,z,t l cc s thc khng m. Chng minh rng:3(x2 + y2 + z2 + t2) + 4

xyzt (x + y + z + t)2

Li gii.

Ta chng minh bt ng t hc tng ng (Tukervici):x4 + y4 + z4 + t4 + 2xyzt x2y2 + y2z2 + z2x2 + t2x2 + t2y2 + z2t2

Khng mt tnh tng qut, gi s t = min {x; y; z; t}Nu t = 0 th ta c:

x4 + y4 + z4 x2y2 + y2z2 + z2x2 (x2 y2)2 + (y2 z2)2 + (z2 x2)2 0

Nu t > 0 ,chun ho t = 1. Ta cn chng minh:x4 + y4 + z4 + 2xyz + 1 x2y2 + y2z2 + z2x2 + x2 + y2 + z2

Mt khc, ta c bt ng thc vi 3 bin dng:x2 + y2 + z2 + 2xyz + 1 2(xy + yz + zx)

nn ta cn ch ra rngx4 + y4 + z4 x2y2 y2z2 z2x2 2(x2 + y2 + z2 xy yz zx)

(x

y)2[(x + y)2

2] + (y

z)2[(y + z)2

2] + (z

x)2[(z + x)2

2]

0

Nh vy, php chng minh hon tt.C 2 trng hp ca ng thc :x = y = z = t hoc x = y = z; t = 0. 2

10.40 Cho a,b,c l cc s thc dng. Chng minh rng:

(a + b + c)

1

a+

1

b+

1

c

9 + 8.(a b)

2 + (b c)2 + (c a)2(a + b + c)2

Li gii.

Bt ng thc cn chng minh tng ng vi:

229


5/116


6/116

1 Cc bt ng thc kinh in

1.1 Bt ng thc gia trung bnh cng v trung bnh nhn (AM-

GM).

Nu a1, a2, . . . , an l cc s thc khng m, tha1 + a2 + . . . + an n na1a2 . . . an.

ng thc xy ra khi v ch khi a1 = a2 = . . . = an.

1.2 Bt ng thc gia trung bnh cng v trung bnh iu ho (AM-

HM).

Nu a1, a2, . . . , an l cc s thc dng, tha1 + a2 + . . . + an

n n

1

a1+ 1

a2+ . . . + 1

an

.

ng thc xy ra khi v ch khi a1 = a2 = . . . = an.

Thc cht y l mt h qu trc tip ca bt ng thc Cauchy - Schwarz. Hai trng hp thngc s dng nht ca bt ng thc ny l khi n = 3 hay n = 4.

Vi n = 3, ta ca + b + c

3 3

1

a +1

b +1

c

,

1

a+

1

b+

1

c 9

a + b + c.

Vi n = 4, ta ca + b + c + d

4 4

1

a+ 1

b+ 1

c+ 1

d

,

1

a+

1

b+

1

c+

1

d 16

a + b + c + d.

1.3 Bt ng thc Cauchy - Schwarz.

Dng s cp ca n c pht biu nh sau:

Nu a1, a2, . . . , an v b1, b2, . . . , bn l cc s thc tu , th(a1b1 + a2b2 + . . . + anbn)2 (a21 + a22 + . . . + a2n)(b1 + b2 + . . . + b2n).

ng thc xy ra khi v ch khia1b1

=a2b2

= . . . =anbn

, trong ta s dng quy c: nu mu

bng 0 th t cng bng 0.

Trong nh gi trn, chn ai =xi

yi,bi =

yi vi xi, yi R; yi > 0, ta thu c bt ng thc

Cauchy - Schwarz dng phn thc:

Nu x1, x2, . . . , xn l cc s thc v y1, y2, . . . , yn, l cc s thc dng, thx21y1

+x22y2

+ . . . +x2nyn (x1 + x2 + . . . + xn)

2

y1 + y2 + . . . + yn.

ng thc xy ra khi v ch khix1y1

=x2y2

= . . . =xnyn

.

6

(xy)3 +

(yz)3 +

(zx)3

=

(xy)3 +

xy

+

(yz)3 +

yz

+

(zx)3 +

zx

xy +yz +zx

2 (xy + yz + zx) xy +yz +zx (xy + yz + zx) + (xy + yz + zx)

3 (xy + yz + zx)

= (xy + yz + zx) + 3 3= (xy + yz + zx)

Bi ton c chng minh xong.ng thc xy ra khi v ch khi x = y = z = 1. 2

10.35 Cho cc s thc dng a,b,c. Chng minh rng:a

2

4a2 + ab + 4b2+

b2

4b2 + bc + 4c2+

c2

4c2 + ca + 4a2 1

Li gii.

Cch 1:S dng bt ng thc Cauchy-Schwarz, ta c : a2

4a2 + ab + 4b2 [ (4a2 + ac + 4c2)] a2

(4a2 + ab + 4b2)(4a2 + ac + c2)

Do ta i chng minh:

[

(4a2 + ac + 4c2)]

a2

(4a2 + ab + 4b2)(4a2 + ac + c2)

1

iu ny tng ng vi:(8

a2 +

ab) [8

a2b2 + abc(a + b + c)] (4a2 + ab + 4b2)Hay:

66a2b2c2 8abc(a3 + b3 + c3) + 8(a3b3 + b3c3 + c3a3) + 3abc [a2(b + c) + b2(a + c) + c2(a + b)]iu ny ng theo AM-GM.Vy bi ton c chng minh xong.ng thc xy ra khi v ch khi x = y = z = 1. 2Cch 2: rng:

(x + 1)2(4x2 + x + 4) 4(x2 + x + 1)2 = x(x 1)2 0Nn ta c:

14x2 + x + 4

12 x + 1

x2 + x + 1.

Thit lp 2 biu thc tng t, ri cng v theo v, ta c:14x2 + x + 4

+1

4y2 + y + 4+

14z2 + z + 4

12

x + 1

x2 + x + 1+

y + 1

y2 + y + 1+

z + 1

z2 + z + 1

Nh vy ta cn chng minh

x + 1

x2 + x + 1+

y + 1

y2 + y + 1+

z + 1

z2 + z + 1 2

Tng ng vix2

x2 + x + 1+

y2

y2 + y + 1+

z2

z2 + z + 1 1.

Bt ng thc ny lun ng theo Vasile Cirtoaje.Bi ton c chng minh xong.ng thc xy ra khi v ch khi x = y = z = 1. 2

227


7/116

3 + ab + bc + ca 6 3abcS nh gi

ab + bc + ca 3abc(a + b + c) = 3abcta a bi ton v chng minh

1 +

abc 2 3abct t = 6

abc 1, ta c bt ng thc trn tng ng vi

t3 + 1 2t2hay l

(1 t)(1 + x x2

) 0Bi ton c chng minh xong. 2Cch 2:Nhn 12(a + b + c) cho hai v, ta s c bt ng thc tng ng l

a2 + b2 + c2 + 5(ab + bc + ca) 6 3abc(a + b + c).Hay vit li l

(a + b + c)2 + 3(ab + bc + ca) 6 3abc(a + b + c).p dng AM-GM hai ln ta s c ngay iu phi chng minh

(a + b + c)2 + 3ab + bc + ca) 2(a + b + c)

3(ab + bc + ca)

6 3

abc(a + b + c).

ng thc xy ra khi v ch khi a = b = c. 2

10.34 Cho cc s thc dng x; y; z tha mn xy + yz + zx = 3.

Chng minh rng: x + 2y2x + 4y + 3z2

+y + 2z

2y + 4z + 3x2+

z + 2x

2z + 4x + 3y2 1

Li gii.

Vx + 2y

2x + 4y + 3z2=

1

3 z

2

3(2x + 4y + 3z2)Nn bt ng thc trn tng ng vi

x2

2y + 4z + 3x2+

y2

2z + 4x + 3y2+

z2

2x + 4y + 3z2 1

3S dng bt ng thc Cauchy-Schwarz, ta c

z22x + 4y + 3z2

x3 +

y3 +

z32

3(x3 + y3 + z3) + 6(xy + yz + zx)Vy, ta cn ch ra rng x3 +y3 +z32

3(x3 + y3 + z3) + 6(xy + yz + zx) 1

3hay l

(xy)3 +

(yz)3 +

(zx)3 xy + yz + zxTht vy, s dng bt ng thc AM-GM, ta c:

226

1.4 Bt ng thc Holder.

Cho xij (i = 1, 2, . . . , m;j = 1, 2, . . . , n) l cc s thc khng m. Khi ta cmi=1

n

j=1

xij

1m

n

j=1

mi=1

x1

m

ij

.

Tng qut hn, nu p1, p2, . . . , pn l cc s thc dng tho mn p1 + p2 + . . . + pn = 1, thmi=1

n

j=1

xij

pi

nj=1

mi=1

xpiij

.

1.5 Bt ng thc Chebyshev.

Cho hai dy s thc a1 a2 . . . an v b1, b2, . . . , bn. Khi

1. Nu b1 b2 . . . bn th nni=1

aibi

ni=1

ai

ni=1

bi

;

2. Nu b1 b2 . . . bn th nni=1

aibi

ni=1

ai

ni=1

bi

.

1.6 Bt ng thc Minkowski.

Cho hai dy s dng a1, a2, . . . , an v b1, b2, . . . , bn. Vi mi r 1, ta cni=1

(ai + bi)r

1r

ni=1

ari

1r

+

ni=1

bri

1r

.

Trng hp r = 2 l trng hp thng c s dng nht ca bt ng thc Minkowski. Khi ta c n

i=1

(ai + bi)2

ni=1

a2i +

ni=1

b2i .

1.7 Bt ng thc Schur.

Cho cc s thc khng m a,b,c. Khi vi mi s thc dng r, ta car(a b)(a c) + br(b a)(b c) + cr(c a)(c b) 0.ng thc xy ra khi v ch khi a = b = c, hoc a = 0 v b = c, hoc cc hon v tng ng.

Hai trng hp thng c s dng nht ca bt ng thc Schur l r = 1 v r = 2.

Vi r = 1, ta c bt ng thc Schur bc baa3 + b3 + c3 + 3abc ab(a + b) + bc(b + c) + ca(c + a),

(a + b + c)3 + 9abc 4(a + b + c)(ab + bc + ca),(b c)2(b + c a) + (c a)2(c + a b) + (a b)2(a + b c) 0,

a2 + b2 + c2 +9abc

a + b + c 2(ab + bc + ca),

7


8/116

a

b + c+

b

c + a+

c

a + b+

4abc

(a + b)(b + c)(c + a) 2.

Vi r = 2, ta thu c bt ng thc Schur bc bna4 + b4 + c4 + abc(a + b + c) ab(a2 + b2) + bc(b2 + c2) + ca(c2 + a2).

1.8 Bt ng thc Vornicu - Schur.

Vi mi s thc a,b,c v x,y,z

0, bt ng thcx(a b)(a b) + y(b c)(b a) + z(c a)(c b) 0

ng nu mt trong cc iu kin sau c tho mn

1. a b c v x y;

2. a b c v z y;

3. a b c v x + z y;

4. a b c 0 v ax by;

5. a b c 0 v cz by;

6. a b c 0 v ax + cz by;

7. x,y,z l di ba cnh ca mt tam gic;

8. x,y,z l bnh phng di ba cnh ca mt tam gic;

9. ax,by,cz l di ba cnh ca mt tam gic;

10. ax,by,cz l bnh phng di ba cnh ca mt tam gic;

11. Tn ti mt hm li t : I R+, trong I l tp xc nh ca a,b,c,sao cho x = t(a), y = t(b), z = t(c).

1.9 Bt ng thc Bernoulli.

Nu 1 hoc 0 th (1 + x)

1 + x,x > 1.Nu 0 1 th (1 + x) 1 + x,x > 1.

1.10 Ba tiu chun SOS thng gp.

Gi s a b c v c: Sa(b c)2 + Sb(c a)2 + Sc(a b)2 0(Sa, Sb, Sc l cc hm chabin a,b,c).Khi bt ng thc ng nu tha mn mt trong cc tiu chun.1.Sb 0, Sb + Sc 0, Sb + Sa 0.2.Vi a,b,c > 0 tha mn Sb 0, Sc 0, a2Sb + b2Sa 0.3.Sb 0, Sc 0, Sa(b c) + Sb(a c) 0

8

1 1

a + b + 1 a + b + c2

(a + b + c)2

(a + b + c)2 2(a + b + c) + a2 + b2 + c2 ab + bc + ca a + b + c

hay ta c iu phi chng minh.ng thc xy ra khi v ch khi a = b = c = 1. 2Cch 2:S dng bt ng thc Cauchy-Schwarz ta c:

2 (1 1a + b + 1 ) = a + ba + b + 1 (a + b + b + c + c + a)

2

(a + b)(a + b + 1) + (b + c)(b + c + 1) + (c + a)(c + a + 1)

a2 + b2 + c2 + ab + bc + ca + a + b + c (a + b + c)2

ab + bc + ca a + b + chay ta c iu phi chng minh.ng thc xy ra khi v ch khi a = b = c = 1. 2Cch 3:Gi s tn ti cc s dng a,b,c sao cho: 1

a + b + 1 1 v a + b + c < ab + bc + ca.

Khi ta c:

1a + b + 1

1 1 >

1 2(ab + bc + ca)(a + b)(a + b + c) + ab + bc + ca

1 > a2 + ab + b2

(a + b)(a + b + c) + ab + bc + ca 3

4

(a + b)2(a + b)(a + b + c) + ab + bc + ca

3(a + b + c)2

[(a + b)(a + b + c) + ab + bc + ca]=

3(a + b + c)2

2(a + b + c)2 + 3(ab + bc + ca) 1

iu cui cng l v l, do bi ton ca ta ng.Php chng minh hon tt.ng thc xy ra khi v ch khi a = b = c = 1. 2

10.33 Cho a,b,c l cc s thc dng. Chng minh rng:a + b + c

3 (a b)

2 + (b c)2 + (c a)212(a + b + c)

+ 3

abc

Li gii.

Cch 1:Ta c bt ng thc cn chng minh tng ng vi

2(a + b + c)2

6(a + b + c) a

2 + b2 + c2 ab bc ca6(a + b + c)

3abcHay:

(a + b + c)2 + 3(ab + bc + ca) 6 3abc(a + b + c)Chun ha cho a + b + c = 3, bt ng thc tr thnh

225


9/116

f(x) =1 1

x ln x

(x 1)2 =g(x)

(x 1)2 .Ly o hm g(x), ta c

g(x) =1

x2 1

x< 0.

Suy rag(x) < lim

x1+g(x) = 0.

Suy raf(x) < 0.

T ta c ngay hm f(x) nghch bin trn (1;+).T suy ra iu phi chng minh. 2.

10.31 Cho a,b,c l cc s thc dng tha mn abc = 1. Chng minh rng:1

a(a + b)+

1

b(b + c)+

1

c(c + a) 3

2

Li gii.

Vit bt ng thc li thnhbc

a + b+

ca

b + c+

ab

c + a 3

2.

Dng bt ng thc hon v, ta c(hoc cng c th chng minh bng phn tch dng M(a b)2 + N(a c)(b c) 0)

bc

a + b+

ca

b + c+

ab

c + a ab

a + b+

bc

b + c+

ca

c + a.

Nh vy (bc cui dng AM-GM)bc

a + b+

ca

b + c+

ab

c + a 1

2

bc

a + b+

ca

b + c+

ab

c + a

+

1

2

ab

a + b+

bc

b + c+

ca

c + a

=

1

2

b(c + a)

a + b+

c(a + b)

b + c+

a(b + c)

c + a

32 3

b(c + a)

a + b c(a + b)

b + c a(b + c)

c + a=

3

2.

Ta c iu phi chng minh.ng thc xy ra khi v ch khi a = b = c = 1. 2

10.32 Cho a,b,c l cc s thc dng tha mn1

a + b + 1

+1

b + c + 1

+1

c + a + 1 1

Chng minh rng:a + b + c ab + bc + ca

Li gii.

Cch 1:S dng bt ng thc Cauchy-Schwarz ta c:

(a + b + 1)(a + b + c2) (a + b + c)2Suy ra:

224

2 Mt s nh gi quen thuc

1 Vi mi s thc a, b, ta lun c2(a2 + b2) (a + b)2

Chng minh. rng

2(a2

+ b2

) (a + b)2 = (a b)2 0,do ta c iu phi chng minh.

ng thc xy ra khi v ch khi a = b. 2

2 Vi mi s thc a,b,c, ta lun ca2 + b2 + c2 ab + bc + ca

Chng minh. rng

a2 + b2 + c2 (ab + bc + ca) = 12

(a b)2 + (b c)2 + (c a)2

0,

do vy ta c iu phi chng minh.

ng thc xy ra khi v ch khi a = b = c. 2

Lu . T nh gi ny ta suy ra(a + b + c)2 3(ab + bc + ca),

v3(a2 + b2 + c2) (a + b + c)2.

3 Vi mi s thc dng a,b,c, ta lun c1

a+

1

b+

1

c 9

a + b + c

Chng minh. y l mt kt qu c cp trn. Li gii c th s dng bt ng thcAM-HM hoc Cauchy - Schwarz. ng thc xy ra khi v ch khi a = b = c. 2

9


10/116

3 Tuyn tp bt ng thc

3.1 Bi 1.1 n bi 1.40

1.1 Cho x,y,z l cc s thc dng tha mn x + y + z = 1. Chng minh rng:8x + 8y + 8z 4x+1 + 4y+1 + 4z+1

Li gii. t a = 2x, b = 2y, c = 2z. Khi iu kin cho c vit li thnha,b,c > 0; abc = 2x+y+z = 64,

v ta cn chng minha3 + b3 + c3 4(a2 + b2 + c2).

rng ta c ng thca3 + 32 6a2 = (a 4)2(a + 2),

t s dng gi thit a > 0 ta suy ra a3 + 32 6a2. Thit lp cc bt ng thc tng t chob v c v cng v theo v cc bt ng thc thu c, ta c

a3 + b3 + c3 + 96 6(a2 + b2 + c2).Nh vy kt thc chng minh ta cn ch ra rng

6(a2 + b2 + c2) 4(a2 + b2 + c2) + 96,hay

2(a2 + b2 + c2) 96.Tuy nhin bt ng thc ny ng theo bt ng thc AM-GM cho ba s:

2(a2

+ b2

+ c2

) 2.33

a2b2c2 = 63

4096 = 96.Nh vy php chng minh n y hon tt. 2

1.2 Cho a,b,c l cc s thc tho mn a 4, b 5, c 6 v a2 + b2 + c2 = 90.Tm gi tr nh nht ca biu thc:

P = a + b + c

Li gii. t a = m + 4, b = n + 5, c = p + 6, khi m,n,p 0 v t gi thit a2 + b2 + c2 = 90ta suy ra

m2 + n2 + p2 + 8m + 10n + 12p = 13.

rng ta c ng thc sau(m + n + p)2 + 12(m + n + p) = (m2 + n2 + p2 + 8m + 10n + 12p) + 2(mn + np + pm + 2m + n).

n y ta s dng cc gi thit cho c

(m + n + p)2

+ 12(m + n + p) 13,t ta suy ra m + n + p 1.Thay m = a 4, n = b 5, p = c 6 ta suy ra a + b + c 10 hay P 16.

Cui cng, vi a = 4, b = 5, c = 7 (tho mn cc iu kin cho) ta c P = 16 nn ta kt lun16 l gi tr nh nht ca biu thc P.

Php chng minh hon tt. 2

1.3 Cho x,y,z l cc s thc tho mn xy + yz + 3zx = 1.Tm gi tr nh nht ca biu thc:

P = x2 + y2 + z2

10

(m2 + n2)(n2 + p2)(p2 + m2)

8m2n2p2

m2 + n2 + p2

mn + np + pm

2.

rng ta c nhn xt sau:Vi x y > 0 v z > 0 th ta c x

y x + z

y + zT nhn xt suy ra

m2 + n2

2mn m

2 + n2 + p2

2mn + p2

m2 + p2

2mp m

2 + n2 + p2

2mp + n2

p2 + n22pn

m2 + n2 + p22pn + m2

Vy ta ch cn chng minh(m2 + n2 + p2)(mn + mp + np)2 (m2 + 2np)(n2 + 2mp)(p2 + 2mn) (m n)2(mp)2(np)2 0

Bt ng thc cui lun ng, vy ta c iu phi chng minh.ng thc xy ra khi v ch khi a = b = c. 2Li gii 2.

Theo Bt ng thc Schur bc 4, ta cabc(a + b + c) (a + b c)c3

Mt khc, theo Bt ng thc Holder ta c:

(a + b c)c3 =

c3

1a + b c2

(a + b + c)3

1a + b c2

Kt hp 2 iu trn, ta suy ra

abc(a + b + c) (a + b + c)3 1

a + b c

2

abc

a + b c +

abc

b + c a +

abc

c + a b a + b + c chnh l iu cn chng minh.ng thc xy ra khi v ch khi a = b = c. 2

10.30 Cho k 1. Chng minh rng:kk (k + 1)k1

Li gii.V k = 1 th bt ng thc tr thnh ng thc nn ta ch cn xt k > 1Ly Logarit Nepe hai v, ta c

k ln k (k 1)ln(k + 1).Hay vit li di dng

ln k

k 1 ln(k + 1)

k.

n y c th thy ngay l ta cn chng minh hm sau nghch binf(x) =

ln x

x 1 vi x > 1.Ly o hm f(x) ta c

223


11/116

x

1 + x2+

y

1 + y2+

z

1 + z2 9

10Bi ton c chng minh xong.ng thc xy ra khi v ch khi x = y = z =

1

3. 2

Cch 2:Ta c:

x

1 + x2 72x

100+

3

50Tht vy, bt ng thc trn tng ng vi 1 bt ng thc lun ng (vi mi x dng) sau:

(4x + 3)(3x

1)2

0

Tng t vi y, z, sau cng v vi v, ta c:x

1 + x2+

y

1 + y2+

z

1 + z2 72

100(x + y + z) +

9

50=

9

10.

Php chng minh hon tt.ng thc xy ra khi v ch khi x = y = z =

1

3. 2

10.28 Cho ba s thc khng m a,b,c tha mn a2 + b2 + c2 = 1. Chng minh rng:a

1 + bc+

b

1 + ca+

c

1 + ab 2

Li gii.

Ta s chng minh rng:(a + b + c)2 2(1 + bc)2

Tht vy, kt hp vi gi thit a2 + b2 + c2 = 1 th bt ng thc trn s tng ng vi:2(ab + bc + ca) 1 + 4bc + 2b2c2 2a(b + c) a2 + (b + c)2 + 2b2c2 (b + c a)2 + 2b2c2 0

Bt ng thc cui lun ng.T ta suy ra:

a

1 + bc a

2

a + b + cTng t 2 biu thc cn li v cng v theo v ta c:

a

1 + bc+

b

1 + ca+

c

1 + ab a

2

a + b + c+

b

2

a + b + c+

c

2

a + b + c=

2

ng thc xy ra khi v ch khi a = b =1

2, c = 0 hoc cc hon v tng ng. 2

10.29 Cho a,b,c l di 3 cnh ca 1 tam gic. Chng minh rng: abca + b c +

abcb + c a +

abcc + a b ab + bc + ca

Li gii.

Li gii 1.

t x = b + c a, y = c + a b, z = a + b c, bt ng thc khi tng ng vi(x + y)(y + z)(z + x)

8

1x

+1y

+1

z

x + y + z.

Bnh phng hai v v quy ng, ta c(x + y)(y + z)(z + x)

xy +

yz +

zx2 8xyz(x + y + z)2.

t tip m =

x, n =

y, p =

z, bt ng thc tr thnh

222

Li gii. t a =9 + 3

17

4v b =

3 +

17

4, khi a = 3b v a + 1 = 2b2 = c =

13 + 3

17

4.

p dng bt ng thc AM-GM ta thu c cc bt ng thc saux2 + b2y2 2bxy,by2 + z2 2byz,

a(z2 + x2) 2azx.n y ta cng v theo v cc bt ng thc thu c c

(a + 1)(x2 + z2) + 2b2y2 2b(xy + yz) + 2azx,hay

c(x2 + y2 + z2) 2b(xy + yz + 3zx).T ta thay cc gi tr ca xy + yz + 3zx, b v c c

P = x2 + y2 + z2

17 32

.

Cui cng, vi x = z =1

4

17v y =

13

17 5134

(tho mn gi thit) th P =

17 32

nn ta

kt lun

17 32

l gi tr nh nht ca biu thc P.


1.4 Cho a,b,c l cc s thc dng tho mn a + b + c. Chng minh rng:a7 + b7

a5 + b5+

b7 + c7

b5 + c5+

c7 + a7

c5 + a5 1

3

Li gii. Trc ht ta c ng thc sau

2(a7 + b7) (a2 + b2)(a5 + b5) = (a b)2(a + b)(a4 + a3b + a2b2 + ab3 + b4),

do vy t gi thit a, b 0 ta suy raa7 + b7

a5 + b5 a

2 + b2

2.

Hon ton tng t ta cng cb7 + c7

b5 + c5 b

2 + c2

2v

c7 + a7

c5 + a5 c

2 + a2

2. n y ta cng v theo

v ba bt ng thc thu c ca7 + b7

a5 + b5+

b7 + c7

b5 + c5+

c7 + a7

c5 + a5 a2 + b2 + c2.

Nh vy kt thc chng minh ta cn ch ra rnga2 + b2 + c2

1

3

.

Tuy nhin bt ng thc trn ng doa2 + b2 + c2 1

3= a2 + b2 + c2 (a + b + c)

2

3=

(a b)2 + (b c)2 + (c a)23

0.Nh vy php chng minh n y hon tt. 2

1.5 Cho a,b,c l cc s thc dng. Chng minh rng:b2c

a3(b + c)+

c2a

b3(c + a)+

a2b

c3(a + b) 1

2(a + b + c)

Li gii. Ta p dng AM-GM cho ba s nh sau:b2c

a3(b + c)+

b + c

4bc+

1

2b 3 3

b2c

a3(b + c).(b + c)

4bc.

1

2b=

3

2a,

11


12/116

t ta suy rab2c

a3(b + c) 3

2a 3

4b 1

4c.

Thit lp hai bt ng thc tng t v cng li, ta suy rab2c

a3(b + c)+

c2a

b3(c + a)+

a2b

c3(a + b)

3

2 3

4 1

4

(a + b + c) =

1

2(a + b + c).


1.6 Cho a,b,c l cc s thc khng m. Chng minh rng:(a + b + c)3 63(a b)(b c)(c a)

Li gii. Bt ng thc ban u mang tnh hon v gia cc bin nn khng mt tnh tng qut,ta gi s a = max{a,b,c}.

Vi a b c th v phi l biu thc khng dng, trong khi v tri l biu thc khng m nnbt ng thc cn chng minh hin nhin ng. Do vy ta xt trng hp a c b. Khi bnhphng hai v ta thu c bt ng thc tng ng sau:

(a + b + c)6 108[(a b)(b c)(c a)]2. rng cc bin khng m, v vi vic sp th t nh trn th

[(a b)(b c)(c a)]2 = [(a b)(c b)(a c)]2 (a c)2a2c2.n y ta p dng bt ng thc AM-GM c

4(a c)2a2c2 = (a c)2.2ac.2ac [(a c)2 + 2ac + 2ac]3

27=

(a + c)6

27,

t ta suy ra

[(a b)(b c)(c a)]2 (a + c)6108

,

v nh vy ta chng minh c bt ng thc ban u v(a + b + c)6 (a + c)6 108[(a b)(b c)(c a)]2.


1.7 Cho a,b,c l cc s thc dng tho mn a + b + c =1

a+

1

b+

1

c. Chng minh rng:

2(a + b + c) a2 + 3 + b2 + 3 + c2 + 3Li gii. D thy bt ng thc cn chng minh tng ng vi mi bt ng thc trong dysau

(2aa2 + 3) + (2bb2 + 3) + (2cc2 + 3) 0,a2 1

2a +a2

+ 3

+b2 1

2b +b2

+ 3

+c2 1

2c +c2

+ 3 0,

a2 1a

2 +

1 +

3

a2

+

b2 1b

2 +

1 +

3

b2

+

c2 1c

2 +

1 +

3

c2

0.

Cc bt ng thc trn u mang tnh i xng gia cc bin nn khng mt tnh tng qut tahon ton c th gi s a b c. Khi khng kh ta suy ra

a2 1a

b2 1

b c

2 1c

v1

2 +

1 +

3

a2

12 +

1 +

3

b2

12 +

1 +

3

b2

.

12

10.25 Cho ba s thc dng a,b,c. Chng minh rng:a2 + b2 + c2 + 2abc + 1 2(ab + bc + ca)

Li gii.

Trong 3 s a,b,c th lun tn ti 2 s nm cng pha so vi 1.Gi s 2 s l a v b.Khi ta c:

c(a 1)(b 1) 0Mt khc, ta thy rng:

a2

+ b2

+ c2

+ 2abc + 1 2ab 2bc 2ca = (a b)2

+ (c 1)2

+ 2c(a 1)(b 1) 0. chnh l iu ta cn chng minh.ng thc xy ra khi v ch khi a = b = c = 1. 2

10.26 Cho ba s thc a,b,c. Chng minh rng:(2 + a2)(2 + b2)(2 + c2) 9(ab + bc + ca)

Li gii.

Ta c: (2 + a2)(2 + b2)(2 + c2) = 4(a2 + b2 + c2) + 2(a2b2 + b2c2 + c2a2) + 8 + a2b2c2

rng ta c cc bt ng thc sau:a2 + b2 + c2

+ a2b2c2 + 2 a2 + b2 + c2 + 3 3a2b2c2

a2 + b2 + c2 + 9abca + b + c

Schur

2 (ab + bc + ca) .v:

3 (a2 + b2 + c2) 3 (ab + bc + ca).2 (a2b2 + b2c2 + c2a2 + 3) 2 (2ab + 2bc + 2ca).

Cng v theo v cc bt ng thc trn ta thu ngay iu phi chng minh.ng thc xy ra khi v ch khi a = b = c. 2

10.27 Cho ba s thc dng x,y,z tha mn x + y + z = 1. Chng minh rng:x

1 + x2+

y

1 + y2+

z

1 + z2 9

10

Li gii.

Cch 1:

S dng bt ng thc AM-GM ta c:x2 + 1 = x2 +

1

9+

1

9+

1

9+

1

9+

1

9+

1

9+

1

9+

1

9+

1

9 10 10

x2

99= 10 5

x

39

Thit lp 2 biu thc tng t, sau cng v theo v, ta c:x

1 + x2+

y

1 + y2+

z

1 + z2 3

10

5

(3x)4 + 5

(3y)4 + 5

(3z)4

.

Mt khc, cng theo bt ng thc AM-GM, ta nhn thy rng:3x + 3x + 3x + 3x + 1 5 5(3x)4.

Tng t vi y, z, v ch x + y + z = 1, ta suy ra:5

(3x)4 + 5

(3y)4 + 5

(3z)4 3T :

221


13/116

(ab + bc + ca)2 3abc(a + b + c)Ta a bt ng thc v chng minh

27 (a2 + b2 + c2)(ab + bc + ca)2Tht vy, theo bt ng thc AM-GM th

(a2 + b2 + c2)(ab + bc + ca)2

a2 + b2 + c2 + ab + bc + ca + ab + bc + ca

3

3= 27

Chng minh hon tt.ng thc xy ra khi v ch khi a = b = c = 1. 2Cch 3. t x = ab + bc + ca. Khi s dng cc bt ng thc quen thuc

(a + b + c)2 3(ab + bc + ca), (ab + bc + ca)2 3abc(a + b + c)Ta c

0 < x 3 v abc x2

9V th

1

a2+

1

b2+

1

c2=

1

a+

1

b+

1

c

2 2

1

ab+

1

bc+

1

ca

=

x2

a2b2c2 6

abcTa s chng minh

x2 6abc (9 2x)a2b2c2Tht vy

V T V P x2 2x2

3 x

2(9 2x)81

=x2(x 3)2(2x + 3)

81 0

Bi ton c chng minh hon ton.ng thc xy ra khi v ch khi a = b = c = 1. 2

10.24 Cho ba s thc dng a,b,c tha mn a2 + b2 + c2 = 3. Chng minh rng:a + b + c a2b2 + b2c2 + c2a2

Li gii.

Cch 1. t x = a2, y = b2, z = c2 th x + y + z = 3 v bt ng thc tr thnhx +

y +

z xy + yz + zx

Tng ng vi2

x +

y +

z

+ x2 + y2 + z2 (x + y + z)2Theo bt ng thc AM-GM th

x +

x + x2 3xV th m

V T 3(x + y + z) = (x + y + z)2Bi ton c chng minh xong.

ng thc xy ra khi v ch khi a = b = c = 1. 2Cch 2. S dng bt ng thc Holder, ta c

x +

y +

z2

(x2 + y2 + z2) (x + y + z)3 = 27Vy ta ch cn chng minh c

(x2 + y2 + z2).(xy + yz + zx)2 27Tht vy, theo bt ng thc AM-GM th

(x2 + y2 + z2)(xy + yz + zx)2

x2 + y2 + z2 + xy + yz + zx + xy + yz + zx

3

3= 27

Chng minh hon tt.ng thc xy ra khi v ch khi a = b = c = 1. 2

220

Nh vy theo bt ng thc Chebyshev ta ca2 1

a

2 +

1 +

3

a2

+

b2 1b

2 +

1 +

3

b2

+

c2 1c

2 +

1 +

3

c2

13

a2 1a

12 +

1 +

3

a2

Nhng theo gi thit ta li c a2 1a

= (a + b + c)

1

a+

1

b+

1

c

= 0

nn ta suy raa2 1

a

2 +

1 +

3

a2

+

b2 1b

2 +

1 +

3

b2

+

c2 1c

2 +

1 +

3

c2

0,

v v vy bt ng thc cho cng ng.


1.8 Cho a,b,c l cc s thc dng tho mn a + b + c = 3. Chng minh rng:ab

c2 + 3+

bca2 + 3

+ca

b2 + 3 3

2

Li gii. Trc ht rng

ab + bc + ca (a + b + c)2

3=

(a b)2 + (b c)2 + (c a)2

6

0,

do t gi thit ta suy ra ab + bc + ca

3. Nh vyab

c2 + 3 ab

c2 + ab + bc + ca=

ab(c + a)(b + c)

.

n y ta p dng bt ng thc AM-GM cab

c2 + 3 1

2

ab

c + a+

ab

b + c

.

Thit lp hai bt ng thc tng t v cng li, ta suy ra dy cc nh gi sauab

c2 + 3+

bca2 + 3

+ca

b2 + 3 1

2

ab

c + a+

bc

c + a

+

bc

a + b+

ca

a + b

+

ca

b + c+

ab

b + c

ab

c2 + 3+

bca2 + 3

+ca

b2 + 3 a + b + c

2t vi lu a + b + c = 3 ta suy ra bt ng thc cho l ng.


1.9 Cho a,b,c l cc s thc dng thay i bt k. Chng minh rng:b + ca

+c + a

b+

a + b

c

2 4(ab + bc + ca)

1

a2+

1

b2+

1

c2

Li gii 1. D thy rng bt ng thc ban u tng ng vi mi bt ng thc trong dysau

[ab(a + b) + bc(b + c) + ca(c + a)]2 4(a + b + c)(a2b2 + b2c2 + c2a2)

a2b2(a + b)2 + 2abc

a(a + b)(a + c) 4

a3b3 + abc

ab(a + b)

Tuy nhin rng

a2b2(a + b)2 4

a3b3

=

a2b2(a b)2 0

13


14/116

v2abc

a(a + b)(a + c)

4

abc

ab(a + b)

= 2abc

a3 + b3 + c3 + 3abc

ab(a + b)

0,

do bt ng thc ban u l ng. Php chng minh n y hon tt. 2

Li gii 2. Bt ng thc ban u mang tnh hon v gia cc bin, nn khng mt tnh tngqut, ta gi s b = max {a,b,c}.

Ta p dng bt ng thc AM-GM nh sau

b + c

a

+c + a

b

+a + b

c 2

= a

b

+b

a

+a

c+ b

c

+c

b

+c

a2

4

a

b

+b

a

+a

cb

c

+c

b

+c

a .Nh vy kt thc chng minh, ta cn ch ra rnga

b+

b

a+

a

c

b

c+

c

b+

c

a

(ab + bc + ca)

1

a2+

1

b2+

1

c2

.

Tuy nhin bng php bin i tng ng ta c(b a)(b c)

ca 0,

l mt nh gi ng do ta gi s b = max {a,b,c}.

Php chng minh n y hon tt. 2

Li gii 3. Bt ng thc ban u mang tnh i xng gia cc bin nn khng mt tnh tngqut, ta gi s b nm gia a v c.

Ta p dng bt ng thc AM-GM nh sau:

4(ab + bc + ca) 1

a2+

1

b2+

1

c2

ab + bc + ca

ca+ ca

1a2

+1

b2+

1

c2

2.

Nh vy kt thc chng minh, ta cn ch ra rngb + c

a+

c + a

b+

a + b

c ab + bc + ca

ca+ ca

1

a2+

1

b2+

1

c2

.

Thc hin php bin i tng ng ta c bt ng thc(a b)(b c)

b2 0,

tuy nhin y li l mt nh gi ng do ta gi s b nm gia a v c.


Nhn xt. Li gii u tin khng mang nhiu ngha lm, v n n thun ch l bin itng ng km theo mt cht tinh trong s dng cc nh gi quen thuc v c bn. y

ta bn thm v hai li gii bng AM-GM.

Ta nhn thy rng pht biu ca bi ton c dng Chng minh rng A2 4BC( y A =

b + c

a+

c + a

b+

a + b

c

2, B = ab + bc + ca v C =

1

a2+

1

b2+

1

c2. Nhn xt ny kh

c bit, n gip ta lin tng n mt nh gi quen thuc sau bng AM-GM:(x + y)2 4xy x, y 0.

Do vy, mt cch t nhin ta ngh ra hai hng gii quyt bi ton trn bng AM-GM:

1. Biu din A = X+ Y, vi X v Y l hai i lng thch hp, sau p dng bt ng thcAM-GM c A2 4XY , t i chng minh XY BC; hoc

14

Li gii.

Cch 1. S dng hng ng thc quen thuc(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

Ta a bt ng thc v dng1

a2+

1

b2+

1

c2+ 2(ab + bc + ca) (a + b + c)2

Theo bt ng thc AM-GM v gi thit ta c1

a2+

1

b2+

1

c2+ 2(ab + bc + ca) 1

ab+

1

bc+

1

ca+ 2

3abc(a + b + c)

= 3 1abc +

abc +

abc 9 = (a + b + c)2.

Php chng minh hon tt.ng thc xy ra khi v ch khi a = b = c = 1. 2Cch 2. S dng bt ng thc AM-GM ta c

1

a2+

1

b2+

1

c2 1

ab+

1

bc+

1

ca=

a + b + c

abc=

3

abcV th ta cn phi chng minh

3

abc a2 + b2 + c2

Hayabc(a2 + b2 + c2) 3

n y ta c hai hng tn cng.Hng 1. Dn binGi s c = min{a,b,c} th 3 = a + b + c 3c, tc c 1 dn n a + b

2 1

tf(a,b,c) = abc(a2 + b2 + c2)

Ta cf(a,b,c) f

a + b

2,

a + b

2, c

= c

ab(a2 + b2) (a + b)

4

8

+

abc2 (bc + ca)

2

4

M theo bt ng thc AM-GM th

(a + b)4 = (a2 + b2 + 2ab)2 8ab(a2 + b2)(bc + ca)2 4bc.ca = 4abc2

nn ta cf(a,b,c) f

a + b

2,

a + b

2, c

cui cng ta ch cn chng minh

fa + b

2, a + b

2, c 3

t x =a + b

2 1, t gii thit ta rt ra c c = 3 2x. Xt

f

a + b

2,

a + b

2, c

3 = (4x5 14x4 + 8x3 9x2 1) = (x 1)2[2x(x 1)(2x 1) + 1] 0

T suy ra iu phi chng minh. 2Hng 2. Dng bt ng thc c inTa vit bt ng thc cn chng minh li nh sau

27 3.abc(a + b + c)(a2 + b2 + c2)S dng bt ng thc quen thuc

219


15/116

10.21 Cho ba s thc khng m a,b,c,d,e. Chng minh rng:a6b + b6c + c6d + d6e + e6a abcde(a2 + b2 + c2 + d2 + e2)

Li gii.

Nu abcde = 0 th bt ng thc hin nhin ng.Vi abcde = 0 ta c bt ng thc tng ng vi

a5

cde+

b5

dea+

c5

eab+

d5

abc+

e5

bcd a2 + b2 + c2 + d2 + e2

S dng bt ng thc AM-GM, ta ca5

cde

+a5

cde

+ c2 + d2 + e2

5

5a5

cde

.a5

cde

.c2.d2.e2 = 5a2

Thc hin tng t cho cc hng t cn li, sau cng v theo v ta c iu phi chng minh.ng thc xy ra khi v ch khi a = b = c = d. 2

10.22 Cho ba s thc dng a,b,c. Chng minh rng:a

a2 + bc+

b

b2 + ac+

c

c2 + ab 3(a + b + c)

2(ab + bc + ca)

Li gii.

Bt ng thc cn chng minh tng ng vi:a2(b + c)

a2 + bc+

b2(c + a)

b2 + ca+

c2(a + b)

c2 + ab+ abc

1a2 + bc

3(a + b + c)2

Ta nhn thy rng bt ng thc ny c suy trc tip t 2 kt qu sau:

1)a2(b + c)

a2 + bc+

b2(c + a)

b2 + ca+

c2(a + b)

c2 + ab a + b + c

2)

2

a2 + bc +

2

b2 + ca +

2

c2 + ab 1

ab +

1

bc +

1

caChng minh 1):t (x; y; z) (a1; b1; c1) , ta chuyn bt ng thc thnh:

x + y

z2 + xy+

y + z

x2 + yz+

z + x

y2 + zx 1

x+

1

y+

1

zKhng mt tnh tng qut, gi s x y z, khi ta c:

V P V T =

1

z x + y

z2 + xy

+

1

x z + x

y2 + zx

+

1

y y + z

x2 + yz

=(z x)(z y)

z3 + xyz+

(y2 x2)(y x)(zx + yz xy)xy(x2 + yz)(y2 + zx)

0.Nh vy, bt ng thc 1) c chng minh.Chng minh 2):

p dng bt ng thc AM-GM: 4a2 + bc

2a

bc 1

ab+

1

acCng v theo v cc bt ng thc tng t ta c bt ng thc 2) c chng minh.Php chng minh hon tt.ng thc xy ra khi v ch khi a = b = c. 2

10.23 Cho ba s thc dng a,b,c tha mn a + b + c = 3. Chng minh rng:1

a2+

1

b2+

1

c2 a2 + b2 + c2

218

2. Biu din BC =B

D.CD, vi D l mt i lng thch hp, sau p dng bt ng thc

AM-GM c 4BC

B

D+ CD

2, t i chng minh A B

D+ CD.

y ta hiu cm t thch hp l nh th no? Lu rng mt trong nhng iu cn trong mi chng minh bt ng thc l cn phi n gin ho bt ng thc cn chng minh. Tac th tm cch gim bc, chun ho iu kin, . . ., nhng tu chung li, ta lun mun bt ngthc cn chng minh tr nn n gin nht c th, t p dng nh nhng cc nh gi

quen thuc hoc bin i tng ng. y ta tm cch thu gn nh gi sau cng theo kiutrit tiu mt lng ng k cc phn t chung, tc l nh gi XY BC hoc A B

D+ CD,

cc i lng X ,Y ,D c chn sao cho hai v ca bt ng thc c nhiu phn t chung ta rt gn. C th:

Hng 1. Trc tin ta vit li A v khai trin tch BC nh sau:

A =b

a+

c

a+

c

b+

a

b+

a

c+

b

c= X+ Y,

BC =a

c+

c

b+

b

a+

a

b+

b

c+

c

a+

ca

b2+

ab

c2+

bc

a2.

rng trong BC c phn tca

b2, nn ta cn c

a

bv

c

b X v Y tng ng:

X =a

b+ . . . , Y =

c

b+ . . .

Mt khc, trong BC c phn ta

b

, m Y cc

b

nn ta cn phn ta

c

trong X:

X =ab

+ac

+ . . . , Y =cb

+ . . .

Tip tc, trong BC c phn tab

c2, nn ta cn c

a

cv

b

c X v Y tng ng:

X =a

b+

a

c+ . . . , Y =

c

b+

b

c+ . . .

Tip tc nh vy ta s tm c hai i lng X, Y chng hn nh sau:X =

a

b+

b

a+

a

c, Y =

b

c+

c

b+

c

a,

v ta c c li gii th hai. Cn lu rng y khng phi l cch chn duy nht.

Hng 2. Xt hiu sau

A BDCD = b + c

a+

c + a

b+

a + b

c ab + bc + ca

DD

1

a2+

1

b2+

1

c2.

rng trong hiu trn th h s ca bin b bng

1

c+

1

a c + a

D,

nh vy tm cch thu gn bt ng thc, ti sao ta khng cho h s ca bin b bng khng?C th, nu chn D = ca th

A BD CD = b + c

a+

c + a

b+

a + b

c ab + bc + ca

ca ca

1

a2+

1

b2+

1

c2

=

(a b)(b c)b2

,

15


16/116

v nh vy ta c li gii th ba.

1.10 Cho a,b,c l cc s thc dng tho mn a + b + c = 1. Tm gi tr ln nht ca biuthc:

P = ab + bc + ca +5

2

(a + b)

ab + (b + c)

bc + (c + a)

ca

Li gii. Trc ht ta p dng bt ng thc AM-GM nh sau:

2(a + b)2 + 2ab =(a + b)2

2

+(a + b)2

2

+(a + b)2

2

+(a + b)2

2

+ 2ab

5

5

ab(a + b)8

8v

(a + b)3 (2

ab)3 = 8(

ab)3,

t kt hp hai bt ng thc ny c

2(a + b)2 + 2ab 5(a + b)

ab.

Thit lp hai bt ng thc tng t v cng li, ta suy ra

5

(a + b)

ab + (b + c)

bc + (c + a)

ca

4(a2 + b2 + c2) + 6(ab + bc + ca)

n y ta cng thm 2(ab + bc + ca) vo mi v c

2(ab + bc + ca) + 5

(a + b)ab + (b + c)bc + (c + a)ca 4(a + b + c)2,t ta suy ra P 2(a + b + c)2 = 2.

Cui cng, vi a = b = c =1

3(tho mn iu kin) th P = 2 nn ta suy ra 2 l gi tr ln nht

ca biu thc P.


1.11 Cho a,b,c l cc s thc dng tho mn1

a+

1

b+

1

c 16(a + b + c). Chng minh rng:

1

(a + b + 2

a + c)3+

1

(b + c + 2

b + a)3+

1

(c + a + 2

c + b)3 8

9

Li gii. Trc ht ta p dng bt ng thc AM-GM nh sau:

a + b +

a + c

2+

a + c

2 3 3

(a + b)(a + c)

2,

t ta suy ra1

(a + b + 2

a + c)3 2

27(a + b)(a + c).

Cng v theo v bt ng thc ny vi hai bt ng thc tng t cho ta

1

(a + b + 2

a + c)3+

1

(b + c + 2

b + a)3+

1

(c + a + 2

c + b)3 4(a + b + c)

27(a + b)(b + c)(c + a).

16

10.19 Cho ba s thc a,b,c tha mn a2 + b2 + c2 = 2. Chng minh rng:|a3 + b3 + c3 abc| 22

Li gii.

t t = ab th ta c t = ab; |t| a2 + b2

2 a

2 + b2 + c2

2= 1

S dng bt ng thc Cauchy-Schwarz ta c:a3 + b3 + c(c2 ab)2 (a + b)2 + c2 (a2 ab + b2)2 + (c2 ab)2

= 2(1 + t)

(c2 ab)2 + (2 c2 ab)2

= 2(1 + t) 2c4 + 2a2b2 + 4 4c2 4ab= 4(1 + t)

t2 2t + 2 + c2(c2 2)

4(t + 1)(t2 2t + 2)Do ta ch cn chng minh

(t + 1)(t2 2t + 2) 2 t2(t 1) 0 .Bt ng thc cui lun ng, php chng minh hon tt.ng thc xy ra khi v ch khi a = b = 0, c = 2. 2

10.20 Cho ba s thc dng a,b,c. Chng minh rng:a4 + b4 + c4

ab + bc + ca+

3abc

a + b + c 2

3(a2 + b2 + c2)

Li gii.

Trc ht, ta c 2 bt ng thc ph sau:Ta c:

a3 + b3 + c3

a + b + c+

1

3(ab + bc + ca) 2

3(a2 + b2 + c2)

3(a3 + b3 + c3) + (a + b + c)(ab + bc + ca) 2(a + b + c)(a2 + b2 + c2) a3 + b3 + c3 + 3abc ab(a + b) + bc(b + c) + ca(c + a)

Lun ng theo Schur bc 3.Ta cng c:

a4 + b4 + c4

ab + bc + ca+

2

3(ab + bc + ca) a2 + b2 + c2

3(a4 + b4 + c4) + 2(ab + bc + ca)2 3(a2 + b2 + c2)(ab + bc + ca) a4 + b4 + c4 + abc(a + b + c) ab(a2 + b2) + bc(b2 + c2) + ca(c2 + a2)

Bt ng thc cui ng do s dng Schur bc 4 v v a4 + b4 + 2a2b2 = (a2 + b2)2

2ab(a2 + b2).

Tr li bi ton, bt ng thc ca bi ton m ta cn chng minh tng ng via4 + b4 + c4

ab + bc + ca+

a3 + b3 + c3

a + b + c a

3 + b3 + c3 3abca + b + c

+2

3(a2 + b2 + c2)

a4 + b4 + c4

ab + bc + ca+

a3 + b3 + c3

a + b + c+ ab + bc + ca 5

3(a2 + b2 + c2)

Tht vt bt ng thc ny ng sau khi cng v theo v ca 2 bt ng thc ph m ta chng minh trn.Php chng minh hon tt.ng thc xy ra khi v ch khi a = b = c. 2

217


17/116

a + b

2 3

abc+

b + c

2 3

abc+

c + a

2 3

abc+

8abc

(a + b)(b + c)(c + a) 4

Hin nhin ng theo bt ng thc AM-GM.ng thc xy ra khi v ch khi a = b = c. 2

10.17 Cho ba s thc dng x,y,z. Chng minh rng:x + y + z

3

xyz+

4xyz

x2y + y2z + z2x + xyz 4

Li gii.

Gi s z l s nm gia 3 s x,y,z. Khi ta c:x(z x)(z y) 0 xz2 + x2y x2z + xyz

S dng nh gi trn v kt hp vi bt ng thc AM-GM, ta c:

x2y + y2z + z2x + xyz z(x + y)2AMGM 427

z + 2.

x + y

2

3=

4(x + y + z)3

27.

S dng kt qu trn, v theo bt ng thc AM-GM, ta c:x + y + z

3

xyz+

4xyz

x2y + y2z + z2x + xyz 3.x + y + z

3 3

xyz+

27xyz

(x + y + z)3AMGM 4.

Bi ton c chng minh xong.ng thc xy ra khi v ch khi x = y = z. 2

10.18 Cho ba s thc khng m a,b,c tha mn a + b + c = 3. Chng minh rng:a + 1b + 1

+ b + 1c + 1

+ c + 1a + 1

253 3

4ab + 4bc + 4ca

Li gii.

S dng bt ng thc AM-GM, ta c3 3

4(ab + bc + ca) 2 + 2 + (ab+ bc + ca) abc+ ab+ bc + ca+ a + b +c +1 = (a +1)(b +1)(c +1) .Vy, ta cn chng minh c

a + 1

b + 1+

b + 1

c + 1+

c + 1

a + 1 25

(a + 1)(b + 1)(c + 1)Hay l

(a + 1)2(c + 1) + (b + 1)2(a + 1) + (c + 1)2(b + 1) 25 ab2 + bc2 + ca2 + (a + b + c)2 + 3(a + b + c) + 3 25 ab2 + bc2 + ca2 4

By gi ta gi s b l s nm gia 3 s a,b,c. Khi ta c:a(b a)(b c) 0 ab2 + a2c abc + a2b

S ng nh gi trn, kt hp vi bt ng thc AM-GM, ta c:ab2 + bc2 + ca2 a2b + bc2 + abc = b(a2 + ac + c2) b(a + c)2

427

b + 2.

a + c

2

3=

4(a + b + c)3

27= 4.

Php chng minh hon tt.ng thc xy ra khi (a,b,c) l mt hon v ca (0, 1, 2). 2

216

Hn na, theo mt kt qu quen thuc, ta li c

(a + b)(b + c)(c + a) 89

(a + b + c)(ab + bc + ca),

do vy

1

(a + b + 2

a + c)3+

1

(b + c + 2

b + a)3+

1

(c + a + 2

c + b)3 1

6(ab + bc + ca).()

n y ta s dng gi thit v nh gi c bn (ab + bc + ca)2 3abc(a + b + c) c

16(a + b + c) 1a

+ 1b

+ 1c 3(a + b + c)

ab + bc + ca,

t suy ra ab + bc + ca 316

. Kt hp vi () ta suy ra

1

(a + b + 2

a + c)3+

1

(b + c + 2

b + a)3+

1

(c + a + 2

c + b)3 8

9.


Nhn xt.

1. C th thy nh gi ban u a + b +

a + c

2+

a + c

2 3 3

(a + b)(a + c)

2chnh l im

mu cht gii quyt bi ton. Thc r a nh gi ny khng kh ngh ti v bi ngmgi cho chng ta phi p dng bt ng thc AM-GM cho ba s.

2. Sau khi nh gi bng AM-GM, ta c th s dng lun gi thit a v bt ng thcthun nht sau:

(a + b + c)

(a + b)(b + c)(c + a) 3(ab + bc + ca)

8abc(a + b + c).

Bt ng thc ny c th c chng minh bng nhiu cch khc nhau.


a+

1

b+

1

c. Chng minh rng:

5(a + b + c) 7 + 8abc

Li gii. Trc ht t gi thit ta c

a + b + c = 1a

+ 1b

+ 1c 9

a + b + c,

t suy ra a + b + c = 3.

Cng t gi thit ta c ab + bc + ca = abc(a + b + c), t y ta suy ra bt ng thc sau l tngng vi bt ng thc cn chng minh

5(a + b + c)2 7(a + b + c) + 8(ab + bc + ca).

rng ta c nh gi c bn sau:

(a + b + c)2 3(ab + bc + ca),

17


18/116

do vy c kt lun cho bi ton ta cn ch ra rng

5(a + b + c)2 7(a + b + c) + 8(a + b + c)2

3,

hay a + b + c 3, l mt nh gi ng do ta chng minh trn.

Do vy bt ng thc ban u c chng minh xong. Bi ton kt thc. 2

1.13 Cho a,b,c l cc s thc dng tho mn1

a+

1

b+

1

c 16(a + b + c). Chng minh rng:

1

2 + a2 +

1

2 + b2 +

1

2 + c2 1Li gii. Bt ng thc cn chng minh tng ng vi

a2

2 + a2+

b2

2 + b2+

c2

2 + c2 1.

p dng bt ng thc Cauchy - Schwarz, ta c

a2

2 + a2+

b2

2 + b2+

c2

2 + c2 (a + b + c)

2

a2 + b2 + c2 + 6.

Nh vy kt thc chng minh ta cn ch ra rng

(a + b + c)2

a2 + b2 + c2 + 6 1.

Thc hin php khai trin tng ng ta c ab + bc + ca 3. Tuy nhin bt ng thc nyng nh vo gi thit ca bi ton. Lu rng t gi thit ta c

ab + bc + ca = abc(a + b + c),

v theo mt nh gi quen thuc th abc(a + b + c) (ab + bc + ca)2

3, t ta suy ra

ab + bc + ca (ab + bc + ca)2

3,

hay ab + bc + ca 3. Php chng minh n y hon tt. 21.14 Cho a,b,c,d l cc s thc dng tho mn a + b + c + d = 1. Tm gi tr nh nht ca

biu thc: P = 1a2 + b2 + c2 + d2

+1

abc+

1bcd

+1

cda+

1dab

Li gii. K hiu

l tng hon v. Trc ht ta s dng AM-GM v gi thit c cc nhgi sau:

abcd

a + b + c + d

4

4=

1

256,

ab + ac + ad + bc + bd + cd 3(a + b + c + d)2

8=

3

8.

Kt hp cc nh gi ny vi bt ng thc Cauchy - Schwarz ta suy ra c cc bt ng thcsau:

18

Gi s z l s nm gia 3 s x,y,z. Khi ta c:x(z x)(z y) 0 xz2 + x2y x2z + xyz

S dng nh gi trn v kt hp vi bt ng thc AM-GM, ta c:

x2y + y2z + z2x + xyz z(x + y)2AMGM 427

z + 2.

x + y

2

3=

4(x + y + z)3

27.

Bi ton c chng minh xong.ng thc xy ra khi v ch khi a = b = c = 1. 2

10.15 Cho ba s thc dng a,b,c. Chng minh rng:2(a3 + b3 + c3)

abc+

9(a + b + c)2

a2 + b2 + c2 33

Li gii.

Li gii 1. Bt ng thc cn chng minh tng ng vi:2 (a3 + b3 + c3 3abc)

abc 9 3(a

2 + b2 + c2) (a + b + c)2a2 + b2 + c2

(a2 + b2 + c2 ab bc ca)

a + b + c

abc 9

a2 + b2 + c2

0

(a2 + b2 + c2 ab bc ca) (a + b + c)(a2 + b2 + c2) 9abc 0y l 1 iu hin nhin ng theo AM-GM, do php chng minh ca ta hon tt.ng thc xy ra khi v ch khi a = b = c. 2

Li gii 2. t p = a + b + c, q = ab + bc + ac,r = abcBt ng thc cn chng minh tng ng vi:

2p3 3pq + 3r

r+ 9

p2

p2 2q 33

2p(p2 3q)

r+ 9

p2

p2 2q 27Ta c r pq

9nn:

2p(p2 3q)

r+ 9

p2

p2 2q 18p2 3q

q+ 9

1 +

2q

p2 2q

Ta s chng minh

18p2 3q

q+ 9

1 +

2q

p2 2q 27

p2

q+

p

p2 2q 4

(p2 3q)2 0Bt ng thc cui ng, vy ta c iu phi chng minh.ng thc xy ra khi v ch khi a = b = c. 2

10.16 Cho ba s thc dng a,b,c. Chng minh rng:a + b + c

3

abc+

8abc

(a + b)(b + c)(c + a) 4

Li gii.

Bt ng thc c th vit li nh sau

215


19/116

P =y

3x+

4x

3y+

z

15x+

16x

15z 4

5S dng bt ng thc AM-GM, ta c:

y

3x+

4x

3y=

1

3(

y

x+

4x

y) 4

3v

z

15x+

16x

15z=

1

15(

z

x+

16x

z) 8

15Do :

P 43

+8

15 4

5=

16

15.

ng thc xy ra 4x = 2y = z a =5b

3 =5c

7 .Vy gi tr nh nht ca P l

16

15. 2

10.14 Cho ba s thc dng a,b,c tha mn abc = 1. Chng minh rng:a

b2 + 3+

bc2 + 3

+c

a2 + 3 3

2

Li gii.

Li gii 1.

Theo bt ng thc AM-GM ta c: ab + bc + ca 3Mt khc, cng theo bt ng thc AM-GM th

ab2 + 3

ab2 + ab + bc + ca

=a

(b + a)(b + c) 2a

a + 2b + c

Thit lp 2 biu thc tng t, ri cng v vi v, ta c:ab2 + 3

+b

c2 + 3+

ca2 + 3

2aa + 2b + c

+2b

b + 2c + a+

2c

c + 2a + bM theo bt ng thc Cauchy-Schwarz ta c:

2a

a + 2b + c+

2b

b + 2c + a+

2c

c + 2a + b 2(a + b + c)

2

a2 + b2 + c2 + 3 (ab + bc + ca)

=2(a + b + c)2

(a + b + c)2 + (ab + bc + ca)

32

.

Do ta c iu phi chng minh.ng thc xy ra khi v ch khi a = b = c = 1. 2

Li gii 2.

V a2b2c2 = 1 nn ta c th thay b (a,b,c) bi

x

y,

z

x,

y

z

.

Khi ta a bt ng thc v dng ng bc lx

3xy + yz+

y3yz + zx

+z

3zx + xy 3

2S dng bt ng thc Holder, ta c x

3xy + yz

2[

x(3xy + yz)] (x + y + z)3

Vy, ta cn chng minh c(x + y + z)3 27

4(x2y + y2z + z2x + xyz)

214

1a2

+ 1

4ab 7

2a2 +

4ab

=49

a2

+ 2

ab 49

1 + 2 38

= 28

7 1

4ab 7 6

24ab

7 364 3

8

= 168.

Mt khc p dng bt ng thc AM-GM cho bn s ta li c abcd

4

1

4abcd 4

11256

= 64.

Kt hp ba bt ng thc va chng minh trn, ta suy ra1

a2 + b2 + c2 + d2+ 2

1ab

+ a

bcd 28 + 168 + 64 = 260 .

Hn na, s dng gi thit a + b + c + d = 1 ta suy ra

P =1

a2 + b2 + c2 + d2+ (a + b + c + d)

1

abc+

1

bcd+

1

cda+

1

dab

=

1

a2 + b2 + c2 + d2+ 2

1ab

+ a

bcd.

Do vy P 260.

Cui cng, vi a = b = c = d =1

4(tho mn iu kin) th P = 260 nn ta suy ra 260 l gi tr

nh nht ca biu thc P.


1.15 Cho x,y,z l cc s thc dng tho mn xyz = 1. Chng minh rng:18

1

x3 + 1+

1

y3 + 1+

1

z3 + 1

(x + y + z)3

Li gii. S dng gi thit, d thy bt ng thc cn chng minh tng ng vi mi bt ngthc trong dy sau:

18

3 x

3

x3 + 1 y

3

y3 + 1 z

3

z3 + 1

(x + y + z)3,

18

x2

x2 + yz+

y2

y2 + zx+

z2

z2 + xy

+ (x + y + z)3 54. ()

p dng bt ng thc Cauchy - Schwarz, ta c

x2

x2 + yz+

y2

y2 + zx+

z2

z2 + xy (x + y + z)

2

x2 + y2 + z2 + xy + yz + zx.

Nh vy nu k hiu V T() l v tri ca bt ng thc () th ta c

V T() 18(x + y + z)2

x2 + y2 + z2 + xy + yz + zx+ (x + y + z)3.

n y ta p dng bt ng thc AM-GM c

V T() 2

18(x + y + z)5

x2 + y2 + z2 + xy + yz + zx.

19


20/116

Nh vy kt thc chng minh, ta cn ch ra rng

(x + y + z)5 812

(x2 + y2 + z2 + xy + yz + zx).

Trc ht ta p dng bt ng thc AM-GM nh sau:

(x + y + z)6 = [(x2 + y2 + z2) + (xy + yz + zx) + (xy + yz + zx)]3 27(x2 + y2 + z2)(xy + yz + zx)2.

Hn na, theo mt kt qu quen thuc ta c (xy + yz + zx)2 3xyz(x + y + z), do

(x + y + z)6 81xyz(x2 + y2 + z2)(x + y + z),

hay (x + y + z)5 81(x2 + y2 + z2) do xyz = 1. Nh vy ta cn ch ra rng

2(x2 + y2 + z2) x2 + y2 + z2 + xy + yz + zx.

Tuy nhin bng php bin i tng ng ta thu c

1

2

(a b)2 + (b c)2 + (c a)2

0,

l mt bt ng thc hin nhin ng. Do vy bt ng thc ban u c chng minh.

Bi ton kt thc. 2

1.16 Cho a,b,c l cc s thc dng tho mn a4 + b4 + c4 = 3. Chng minh rng:a2

b + c+

b2

c + a+

c2

a + b 3

2

Li gii. Ta s i chng minh

a2

b + c+

b2

c + a+

c2

a + b 3

2

4

a4 + b4 + c4

3,

t s dng gi thit suy ra kt lun cho bi ton. Tht vy, p dng bt ng thc Holder,ta c

a2

b + c+

b2

c + a+

c2

a + b

2[a2(b + c)2 + b2(c + a)2 + c2(a + b)2] (a2 + b2 + c2)3.

Hn na, theo mt kt qu quen thuc, ta c

2(a2 + b2) (a + b)2,

t y ta thit lp hai nh gi tng t ca2

b + c+

b2

c + a+

c2

a + b

2[2a2(b2 + c2) + 2b2(c2 + a2) + 2c2(a2 + b2)] (a2 + b2 + c2)3,

haya2

b + c+

b2

c + a+

c2

a + b 1

2

(a2 + b2 + c2)3

a2b2 + b2c2 + c2a2.

20

Vy chng minh bi ton ta cn chng minh bt ng thc mnh hn sau y :a2

b+

b2

c+

c2

a+ a + b + c 2 a2 ab + b2 +b2 bc + c2 +c2 ca + a2

hay l: a2 ab + b2

b+ b

+

b2 bc + c2

c+ c

+

c2 ca + a2

a+ a

2

a2 ab + b2 +

b2 bc + c2 +

c2 ca + a2

Bt ng thc cui cng hin hin ng theo bt ng thc AM-GM nn ta c iu phi chngminh.

ng thc xy ra khi v ch khi a = b = c.2

10.12 Cho ba s thc a,b,c (0;1). Chng minh rng:(a a2)(b b2)(c c2) (a bc)(b ca)(c ab)

Li gii.

Ta c:(a a2)(b b2)(c c2) = abc abc2 + abc.(ab + bc + ca) abc.(a + b + c)

v:(a bc)(b ca)(c ab) = abc abc2 + abc.(a2 + b2 + c2) (a2b2 + b2c2 + c2a2)

Khi , bt ng thc tng ng vi: abc.(a2 + b2 + c2) (a2b2 + b2c2 + c2a2) abc.(a + b + c) abc.(a + b + c) abc.(a2 + b2 + c2) abc.(ab + bc + ca) (a2b2 + b2c2 + c2a2) abc.(a + b + c) abc.[(a b)2 + (b c)2 + (c a)2] b2.(c a)2 + c2.(a b)2 + a2.(b c)2 Sa.(b c)2 + Sb.(c a)2 + Sc.(a b)2 0 (1).

Vi: Sa = a2 abc; Sb = b2 abc; Sc = c2 abc.M:

Sa + Sb + Sc = a2 + b2 + c2 3a2b2c2 3.(abc) 23 3(abc)2 0.(do: abc (0, 1)).

V:Sa.Sb + Sb.Sc + Sc.Sa =

a2b2 + 3a2b2c2 2abc.(ab + bc + ca) 0.

(v: a2b2 + a2b2c2 2a2b2c; b2c2 + a2b2c2 2b2c2a; c2a2 + a2b2c2 2c2a2b)Nn theo nh l S.O.S ta c Bt ng thc (1) ng.Do ta c iu phi chng minh.ng thc xy ra khi v ch khi a = b = c. 2

10.13 Cho ba s thc dng a,b,c. Tm gi tr nh nht ca:P =

a + b

a + b + c+

b + c

b + c + 4a+

c + a

c + a + 16b

Li gii.

t

a + b + c = x

b + c + 4a = y

c + a + 16b = z

a =y x

3

b =z x

15

c =21x 5y z

15

a + b =5y + z 6x

15

b + c =4x y

3

c + a =16x z

15Khi ta c:

213


21/116

(a + b)

c2 (a b)2

2c

ab a + b

2(a2 + b2)

2abc2

c2 (a b)2 a2 + b2

2ab(a b)2

c2 (a b)2 (a b)2

a2 + b2 c2.Bt ng thc cui lun ng do ABC l tam gic nhn.

Vy:cos

AB

2

a + b

2(a2 + b2)

Thit lp 2 biu thc tng t v cng v vi v, ta c:

cos

AB

2

+ cos

B C

2

+ cos

CA

2

2

2

a + ba2 + b2

+b + cb2 + c2

+c + ac2 + a2

.

chnh l iu m ta cn chng minh. 2

10.11 Cho ba s thc dng a,b,c. Chng minh rng:a2

b+

b2

c+

c2

a a2 ab + b2 +b2 bc + c2 +c2 + a2 ac

Li gii.

Cch 1:S dng Bt ng thc AM-GM ta c:

a2 ab + b2b

+ b 2a2 ab + b2Tung t ta cng c

b2 bc + c2c

+ c 2b2 bc + c2c2 ca + a2

a+ a 2c2 ca + a2

Cng v vi v ba bt ng thc trn, ta c:a2

b+

b2

c+

c2

a 2 a2 ab + b2 +b2 bc + c2 +c2 + a2 ac a b c

Li c theo Bt ng thc Mikowsyki, ta c:

a2

ab + b2 +

b2

bc + c2 +

c2 + a2

ac

a b

2

+ b

c

2

+ c

a

22

+3

4

(a + b + c)2

= a + b + c.

T ta c iu phi chng minh.ng thc xy ra khi v ch khi a = b = c. 2Cch 2:Ta vit bt ng thc cn chng minh li nh sau:

2

a2

b+

b2

c+

c2

a

2 a2 ab + b2 +b2 bc + c2 +c2 ca + a2

S dng bt ng thc Cauchy-Schwarz, ta c:a2

b+

b2

c+

c2

a (a + b + c)

2

a + b + c= a + b + c

212

Nh vy kt thc chng minh ta cn ch ra rng(a2 + b2 + c2)3

a2b2 + b2c2 + c2a2 3 4

a4 + b4 + c4

3.

Thc hin php bin i tng ng ta thu c

(a2 + b2 + c2)6 27(a4 + b4 + c4)(a2b2 + b2c2 + c2a2)2.

Tuy nhin bt ng thc trn ng nu ta p dng bt ng thc AM-GM nh sau:

(a2 + b2 + c2)6 = [(a4 + b4 + c4) + (a2b2 + b2c2 + c2a2) + (a2b2 + b2c2 + c2a2)]3

27(a4 + b4 + c4)(a2b2 + b2c2 + c2a2)2


1.17 Cho a,b,c l cc s thc dng tho mn a + b + c = 3. Chng minh rng:a

a + b + 1+

b

b + c + 1+

c

c + a + 1 1

Li gii. S dng gi thit, ta thy rng cc bt ng thc sau l tng ng vi bt ng thccn chng minh

a

4

c+

b

4

a+

c

4

b 1,

a(4 a)(4 b) + b(4 b)(4 c) + c(4 c)(4 a) (4 a)(4 b)(4 c),a2b + b2c + c2a + abc 4.

Bt ng thc trn mang tnh hon v gia cc bin nn khng mt tnh tng qut, ta gi s cnm gia a v b. Khi

a(a c)(b c) 0.Thc hin php khai trin ta c a2b +c2a a2c+ abc. T y ta cng thm i lng (b2c +abc)vo hai v c

a2b + b2c + c2a + abc a2c + b2c + 2abc = c(a + b)2.

n y ta p dng AM-GM nh sau:

c(a + b)2 =1

22c(a + b)(a + b) (2c + a + b + a + b)

3

2.27= 4,

t suy ra a2b + b2c + c2a + abc 4, tc l bt ng thc ban u c chng minh.

Bi ton hon tt. 2

1.18 Cho a,b,c l cc s thc khng m tho mn a + b + c = 1. Chng minh rng:25

27 (1 4ab)2 + (1 4bc)2 + (1 4ca)2 3

Li gii.

21


22/116

1. Chng minh (1 4ab)2 + (1 4bc)2 + (1 4ca)2 3.

Trc ht ta c1 = a + b + c a + b 2

ab,

t suy ra 1 4ab. n y ta s dng gi thit cc bin khng m c0 1 4ab 1,

t m (1 4ab)2 1. Thit lp hai nh gi tng t v cng li ta c ngay iu phichng minh.

2. Chng minh (1 4ab)2 + (1 4bc)2 + (1 4ca)2 2527

.

D thy bt ng thc trn tng ng vi mi bt ng thc trong dy sau:

3 8(ab + bc + ca) + 16(a2b2 + b2c2 + c2a2) 2527

,

ab + bc + ca 2(a2b2 + b2c2 + c2a2) 727

.

rng ta c ng thc sau

ab 2a2b2 59

ab 1

9

7

81= 2

ab 1

9

2,

do ta suy ra ab

2a2b2

5

9 ab1

9+7

81. n y ta thit lp hai nh gi tng t

v cng li c

ab + bc + ca 2(a2b2 + b2c2 + c2a2) 59

ab + bc + ca 1

3

7

27.

Hn na, theo mt kt qu quen thuc ta c ab + bc + ca (a + b + c)2

3=

1

3, do vy ta suy

raab + bc + ca 2(a2b2 + b2c2 + c2a2) 7

27,

tc l bt ng thc ban u c chng minh.

Tm li ta chng minh c25

27 (1 4ab)2 + (1 4bc)2 + (1 4ca)2 3. Php chng minh

hon tt. 2

1.18 Cho x,y,z l cc s thc dng tho mn xy + yz + zx = 1. Chng minh rng:1

1 + xy + z2+

1

1 + yz + x2+

1

1 + zx + y2 9

5

Li gii. t x =1

a, y =

1

b, z =

1

c. Khi s dng gi thit xy + yz + zx = 1, ta thy rng

1

1 + xy + z2=

xy + yz + zx

x2 + xy + xz + 2yz=

1

ab+

1

bc+

1

ca1

a2+

1

ab+

1

ac+

2

bc

=a(a + b + c)

2a2 + ab + bc + ca,

22

2a2 +

2

a + 1+ b4 +

2b2 +

2

b + 1+ c4 +

2c2 +

2

c + 1+ a4

2(a + b + c)2 + 2

1

a + 1+

1b + 1

+1

c + 1

2+ (a + b + c)2

=

27 + 2

1

a + 1+

1b + 1

+1

c + 1

2

27 + 2

3

6(a + 1)(b + 1)(c + 1)2

=

27 +

183

(a + 1)(b + 1)(c + 1)

27 +54

a + 1 + b + 1 + c + 1

= 6.

Nh vy ta c iu phi chng minh.ng thc xy ra khi v ch khi a = b = c = 1. 2

10.10 Cho tam gic ABC nhn. Chng minh rng

cos

AB

2

+ cos

B C

2

+ cos

C A

2

2

2

a + ba2 + b2

+b + cb2 + c2

+c + ac2 + a2

Li gii.Bng 1 s php bin i lng gic trong tam gic, ta c:

cos

AB

2

=

cos

AB

2

cos

A + B

2

sin

C

2

=cos A + cos B

2

1 cos C

2

=

b2 + c2 a22bc

+a2 + c2 b2

2ac

2

1 a2 + b2 c2

2ab2

=

(a + b)(c2 (a b)2)abc

2

c2 (a b)2

ab

=(a + b)

c2 (a b)2

2c

abTa s chng minh

211


23/116

P

(x + y + z)2 +

1

x+

1

y+

1

z

2+

2

x+

2

y+

2

z

2

(x + y + z)2 +81

(x + y + z)2+

324

(x + y + z)2

=

(x + y + z)2 +

81

16(x + y + z)2

+

6399

16(x + y + z)2

2.9

4+

6399.4

16.9

= 272

.

ng thc xy ra khi v ch khi x = y = z =1

2.

Vy gi tr nh nht ca P l27

2khi x = y = z =

1

2. 2

10.8 Cho ba s thc dng a,b,cChng minh rng:

ab

a + 9b + 6c+

bc

b + 9c + 6a+

ca

c + 9a + 6b a + b + c

16

Li gii.

S dng bt ng thc Cauchy-Schwarz, ta c:16

a + 9b + 6c

=(1 + 3)2

(3c + a) + 3(3b + c)

1

3c + a

+3

3b + c

.

T ta suy raab

a + 9b + 6c 1

16

ab

3c + a+

3ab

3b + c

Chng minh tng t vi 2 biu thc cn li, sau cng v vi v, ta c:

ab

a + 9b + 6c+

bc

b + 9c + 6a+

ca

c + 9a + 6b

116

ab

3c + a+

3ab

3b + c+

bc

3a + b+

3bc

3c + a+

ca

3b + c+

3ca

3a + b

=

a + b + c

16.

Bi ton c chng minh xong.ng thc xy ra khi v ch khi a = b = c. 2

10.9 Cho a; b; c dng v a + b + c = 3. Chng minh rng2a2 +

2

a + 1+ b4 +

2b2 +

2

b + 1+ c4 +

2c2 +

2

c + 1+ a4 6

Li gii.

S dng bt ng thc Minkowski, kt hp vi bt ng thc AM-GM v gi thit, ta c

210

do bt ng thc cho tng ng vi a2a2 + ab + bc + ca

95(a + b + c)

.

Nhn c hai v ca bt ng thc ny vi ab + bc + ca v ch rng

a(ab + bc + ca)

2a2 + ab + bc + ca= a 2a

3

2a2 + ab + bc + ca,

ta c

2 a3

2a2 + ab + bc + ca +

9(ab + bc + ca)

5(a + b + c) a + b + c.p dng bt ng thc Cauchy - Schwarz, ta c

a32a2 + ab + bc + ca

a22

a(2a2 + ab + bc + ca)

=

a22

6abc +

a

2

a2

ab .

(1)

Mt khc, t bt ng thc c bn (ab + bc + ca)2 3abc(a + b + c), ta li c

3abc (ab + bc + ca)2

a + b + c . (2)

Kt hp (1) v (2), ta suy ra

a32a2 + ab + bc + ca

a22

a

2

ab + bc + ca2

+

a2

2

a2

ab .

=

a2

a

2

a2 + 3

ab.

Cui cng ta ch cn chng minh

2(a2 + b2 + c2)(a + b + c)

2(a2 + b2 + c2) + 3(ab + bc + ca) +

9(ab + bc + ca)

5(a + b + c) a + b + c.Sau khi khai trin v rt gn, ta c bt ng thc hin nhin ng

(ab + bc + ca)(a2 + b2 + c2 ab bc ca) 0.

Bi ton c chng minh xong. 2


a+

1

b+

1

c. Chng minh rng:

(b + c a)(c + a b)(a + b c) 1

Li gii 1. Bt ng thc cn chng minh mang tnh i xng gia cc bin, do khng mt

23


24/116

tnh tng qut, ta gi s a b c. Khi a + b c 0 v c + a b 0.

Nu b + ca < 0 th bt ng thc hin nhin ng do (b + ca)(c + ab)(a + bc) 0 < 1. Do ta ch cn gii quyt bi ton trong trng hp b + ca 0. Lc ny ta t x = b + ca, y =c + a b, z = a + b c. Khi ta vit li iu kin nh sau

x,y,z 0; x + y + z = 2x + y

+2

y + z+

2

z + x,

v ta cn chng minh

xyz 1.Ta s gii quyt bi ton bng phng php phn chng. Tht vy, gi s rng xyz > 1. Khi s dng bt ng thc AM-GM, ta suy ra

x + y + z =2

x + y+

2

y + z+

2

z + x 1

xy+

1yz

+1zx

,

hay

x +

y +

z xyz(x + y + z). Hn na, ta cng c xyz > 1 nn

x +

y +

z > x + y + z.

Tuy nhin theo bt ng thc AM-GM, ta li c

x x + 12

. Ta thit lp thm hai nh gitng t na c

x + y + z + 3

2

x +

y +

z > x + y + z,

hay x + y + z < 3. Nhng y l mt nh gi sai v theo mt kt qu quen thuc, ta c

x + y + z =2

x + y+

2

y + z+

2

z + x 9

x + y + z,

dn ti x + y + z 3. Mu thun ny chng t iu gi s ban u l sai, do vy xyz 1.


Li gii 2. Bt ng thc cn chng minh mang tnh i xng gia cc bin, do khng mttnh tng qut, ta gi s a b c. Khi a + b c 0 v c + a b 0.

Nu b + ca < 0 th bt ng thc hin nhin ng do (b + ca)(c + ab)(a + bc) 0 < 1. Do

ta ch cn gii quyt bi ton trong trng hp b + ca 0. Lc ny ta t x = b + ca, y =c + a b, z = a + b c. Khi ta vit li iu kin nh sau

x,y,z 0; x + y + z = 2x + y

+2

y + z+

2

z + x,

v ta cn chng minhxyz 1.

Ta s gii quyt bi ton bng phng php phn chng. Tht vy, gi s rng xyz > 1. Khi ,t gi thit, ta suy ra

(x + y + z)2(xy + yz + zx) = 2(x + y + z)2 + 2(xy + yz + zx) + xyz(x + y + z). ()

24

Du ng thc xy ra khi a = b = c hoc trong b 3 s sau v cc hon v (a,b,c) =k

sin2 47

, sin2 27

, sin2 7

. 2

10.6 Cho ba s thc dng x,y,z tha mnxy + yz + zx = 3.

Chng minh rng:1

xyz+

4

(x + y)(y + z)(z + x) 3

2

Li gii.

Theo Bt ng thc AM-GM ta c:1xyz

+4

(x + y)(y + z)(z + x)=

1

2xyz+

1

2xyz+

4

(x + y)(y + z)(z + x)

12xyz

+2

2xyz(x + y)(y + z)(z + x)

.

Mt khc, cng theo Bt ng thc AM-GM ta thy rng:

xyz =

xy.yz.zx

xy + yz + zx

3

3= 1

vxyz(x + y)(y + z)(z + x) = (xz + yz)(yx + zx)(zy + xy)

xz + yz + yx + zx + zy + xy

3

3= 8.

T 1

xyz+

4

(x + y)(y + z)(z + x) 1

2xyz+

2

2xyz(x + y)(y + z)(z + x)

.

12

+2

28

=3

2.

Bi ton c chng minh xong.ng thc xy ra khi v ch khi x = y = z = 1. 2

10.7 Cho ba s thc dng x,y,z tha mnx + y + z 3

2.

Tm gi tr nh nht ca:P =

x2 +

1

y2+

4

z2+

y2 +

1

z2+

4

x2+

z2 +

1

x2+

4

y2

Li gii.

S dng ln lt bt ng t hc Mincopski, Cauchy-Schwarz, AM-GM v kt hp vi gi thit, tac:

209


25/116

ng thc xy ra khi v ch khi a = b = c = 13. 2

10.5 Cho ba s thc dng a,b,c. Chng minh rng:(a2 + b2 + c2)2 3(a3b + b3c + c3a)

Li gii.

Li gii 1.

Bt ng thc cn chng minh tng ng vi:2(a2 + b2 + c2)2 6(a3b + b3c + c3a) 0

(a2 2ab + bc c2 + ca)2

0

iu ny hin nhin ng. Bt ng thc c chng minh.Du ng thc xy ra khi a = b = c hoc trong b 3 s sau v cc hon v (a,b,c) =k

sin2 47

, sin2 27

, sin2 7

. 2

Li gii 2.

Ta bit vi mi s thc x,y,z th(x + y + z)2 3(xy + yz + zx)

chnx = a2 + bc ab, y = b2 + ca bc, z = c2 + ab ca

ta thu c[

(a2 + bc ab)]2 3(a2 bc ab)(b2 + ca bc)Mt khc, ta thy rng

(a2 + bc ab) = a2 + b2 + c2

v (a2 bc ab)(b2 + ca bc) = a3b + b3c + c3a

Nn ta c(a2 + b2 + c2)2 3(a3b + b3c + c3a).

Php chng minh hon tt.Du ng thc xy ra khi a = b = c hoc trong b 3 s sau v cc hon v (a,b,c) =k

sin2 47

, sin2 27

, sin2 7

. 2

Li gii 3.

Bng cch xt hiu hai v ca bt ng thc, ta c

a2

2

3

a2b

=1

2a2

34

+ 54

.ab + 5

2ca +

54 1

4

.b2 +

34 5

4

.bc

14

+ 54

.c22

=1

3.a2

1

2+

15

6

.ab +

15

3.ca +

15

6 1

6

.b2 +

1

2

15

6

.bc

1

6+

15

6

.c2

2

=1

4.a2

3

8+

29

8.ab

+

29

4.ca +

29

8 1

8

.b2 +

3

8

29

8

.bc

1

8+

29

8

c2

2 0.

T suy ra iu phi chng minh.

208

Tuy nhin, theo bt ng thc AM-GM v theo iu gi s trn, ta c cc nh gi

xy + yz + zx 3 3

x2y2z2 > 3,

x + y + z 3 3xyz > 3,do vy ta suy ra

2(x + y + z)2(xy + yz + zx)

3> 2(x + y + z)2,

2(x + y + z)2(xy + yz + zx)

9> 2(xy + yz + zx),

(x + y + z)2(xy + yz + zx)9

> xyz(x + y + z).

Cng v theo v cc nh gi trn li, ta c

(x + y + z)2(xy + yz + zx) > 2(x + y + z)2 + 2(xy + yz + zx) + xyz(x + y + z),

tri vi (). Mu thun ny chng t iu gi s ban u l sai, do vy xyz 1.


1.20 Cho a,b,c l cc s thc dng tho mn a + b + c = 3. Chng minh rng:1

5a2 + ab + bc+

1

5b2 + bc + ca+

1

5c2 + ca + ab 3

7

Li gii. p dng bt ng thc Cauchy - Schwarz, ta c

1

5a2 + ab + bc+

1

5b2 + bc + ca+

1

5c2 + ca + ab=cyc

(b + c)2

(b + c)2(5a2 + ab + bc)

4(a + b + c)2

cyc

(b + c)2(5a2 + ab + bc).

Theo , ta cn chng minh rng

4(a + b + c)2cyc

(b + c)2(5a2 + ab + bc) 3

7.

S dng gi thit a + b + c = 3, ta thy rng bt ng thc trn tng ng vi

28(a + b + c)4 27cyc

(b + c)2(5a2 + ab + bc)

.

Sau khi khai trin v rt gn, ta c

28

a4 + 58cyc

a3b + 85cyc

ab3 156

a2b2 + 15abc(a + b + c).

chng minh bt ng thc ny, trc ht ta ch n cc nh gi c bn sau ( thu c bngbt ng thc AM-GM):

cyc

a3b +cyc

ab3 2

a2b2,

25


26/116

a4 +

cyc

ab3 cyc

a3b +cyc

ab3 2

a2b2,

a4

a2b2 abc(a + b + c).

T ta suy ra58cyc

a3b + 58cyc

ab3 116

a2b2,

27

a4 + 27cyc

ab3 54

a2b2,

a4 + 14 a2b2 15abc(a + b + c).Cng v theo v cc nh gi trn, ta thu c bt ng thc cn chng minh.

Bi ton kt thc. 2

1.21 Cho a,b,c l cc s thc dng thay i bt k. Chng minh rng:b + c

2a2 + bc+

c + a

2b2 + ca+

a + b

2c2 + ab 6

a + b + c

Li gii. Nhn c hai v ca bt ng thc cho 4(a + b + c), ta c4(b + c)(a + b + c)

2a2 + bc+

4(c + a)(a + b + c)

2b2 + ca+

4(a + b)(a + b + c)

2c2 + ab 24.

Do4(b + c)(a + b + c)

2a2 + bc=

(a + 2b + 2c)2

2a2 + bc a2

2a2 + bcnn ta c (a + 2b + 2c)22a2 + bc

24 + a2

2a2 + bc.

Bt ng thc ny c suy ra bng cch cng hai bt ng thca2

2a2 + bc+

b2

2b2 + ca+

c2

2c2 + ab 1,

(a + 2b + 2c)2

2a2 + bc+

(b + 2c + 2a)2

2b2 + ca+

(c + 2c + 2b)2

2c2 + ab 25.

Doa2

2a2 + bc=

1

2 bc

2(2a2 + bc)nn bt ng thc th nht tng ng vi

bc

2a2 + bc+

ca

2b2 + ca+

ab

2c2 + ab 1,

ng v theo bt ng thc Cauchy - Schwarz bc

2a2 + bc

bc2

bc(2a2 + bc)

= 1.

By gi ta s chng minh bt ng thc th hai. y l bt ng thc i xng nn khng mttnh tng qut, ta gi s c = min{a,b,c}. t t = b + c

2, ta s chng minh

(a + 2b + 2c)2

2a2 + bc+

(b + 2c + 2a)2

2b2 + ca 2(3t + 2c)

2

2t2 + tc. ()

S dng bt ng thc Cauchy - Schwarz, ta c(a + 2b + 2c)2

2a2 + bc+

(b + 2c + 2a)2

2b2 + ca [b(a + 2b + 2c) + a(b + 2c + 2a)]

2

b2(2a2 + bc) + a2(2b2 + ca)=

2(4t2 ab + 2tc)22a2b2 3abtc + 4t3c .

V tc ab t2 nn

26

Chng minh 1): Ta c:x6y6(x4 + y4) x4y4(x4 + y4 + 2x2y2)2

8=

(xy(x2 + y2))4

8 2

T ta c iu phi chng minh.ng thc ch xy ra khi x = y = 1. 2

10.3 Cho ba s thc khng m a,b,c tha mna2 + b2 + c2 = 3.

Chng minh rng:a3b2 + b3c2 + c3a2 3

Li gii.Gi (x; y; z) l mt hon v ca (a; b; c) sao cho x y z.Khi theo bt ng thc hon v ta c:

a3b2 + b3c2 + c3a2 x3y2 + x2yz2 + z3y2do 2 dy x; y; z v x2y2; z2x2; y2z2 n iu cng chiu.Nh vy ta s chng minh:

x3y2 + yx2z2 + z3y2 3p dng bt ng thc AM-GM:

2x3y2 y(x4 + x2y2)2z3y2 y(z4 + z2y2)

Ch cn cn chng minh (vi ch x2 + y2 + z2 = a2 + b2 + c2 = 3):y(x4 + x2y2) + y(z4 + z2y2) + 2yz2x2 6

y(x2

+ z2

)(x2

+ y2

+ z2

) 6 y(3 y2) 2

(y 1)2(y + 2) 0Bt ng thc cui lun ng, nn ta c iu phi chng minh.ng thc xy ra khi v ch khi x = y = z hay a = b = c = 1. 2

10.4 Cho ba s thc dng a,b,c tha mn a + b + c = 1. Tm gi tr ln nht caab

c + ab+

bc

a + bc+

ca

b + ca 3

2

Li gii.

Vi gi thit a + b + c = 1 th:

ab

c + ab

= ab

1 a b + ab=

ab

(1 a)(1 b)=

ab

(b + c)(c + a)Theo bt ng thc AM-GM ta c:ab

c + ab=

ab

(b + c)(c + a) 1

2

a

c + a+

b

b + c

Tng t ta cng c:

bc

a + bc 1

2

b

a + b+

c

c + a

;

ca

b + ca 1

2

c

b + c+

a

a + b

Cng v vi v ca 3 bt ng thc trn ta c:

ab

c + ab+

bc

a + bc+

ca

b + ca 1

2

a

c + a+

b

b + c+

b

b + a+

c

c + b+

c

c + a+

a

a + b

=

3

2 chnh l iu cn chng minh.

207


27/116

Do vy ta c xyz + 9 z(x + y) + 3(x + y)Ta cn chng minh z(x + y) + 3(x + y) 4(x + y + z)Hay z

x + y

x + y 4Gi s ngc li z

1

x+

1

y+

x + y 4x + y

4x + y

+x + y 4

x + y= 1(v l)

Do vy ta c iu phi chng minh. 2

3.10 Bi 10.1 n bi 10.40

10.1 Cho ba s thc khng m a; b; c . Chng minh rng:a2 + b2 + c2

ab + bc + ca+

4abc

a2b + b2c + c2a + abc 2

Li gii.

Khng mt tnh tng qut, gi s b l s nm gia a v c. Khi ta c:a2b + b2c + c2a + abc = b(c + a)2 + c(a b)(c b) b(c + a)2

Ta cn chng minh bt ng thc:

a2 + b2 + c2

ab + bc + ca+

4ca

(c + a)2 2

a2 + b2 + c2

ab + bc + ca 2(c

2 + a2)

(c + a)2

a2 + b2 + c2

c2 + a2 2(ab + bc + ca

(c + a)2

b2

c2 + a2+

a2 + c2

(c + a)2 d 2b

c + a

(ab + bc a2 c2)2 0

Bt ng thc ny hin nhin ng . ng thc ch xy ra khi a = b = c. 2

10.2 Cho hai s thc dng x; y tha mnx + y = 2. Chng minh rng:

(x4

+ y4

)(x2

+ y2

)4

32

x10y10

Li gii.

Cch gi 1:

(x4 + y4)(x2 + y2)4x10y10 =1

8(x4 + y4)(x3y + xy3)4.(2x2y2)3

1

8.((x + y)4)8

88= 32 (Theo c si 8

s)Cch gii 2:Bt ng thc trn thc cht l h qu ca 2 bt ng thc sau: 1) x6y6(x4 + y4) 22) xy(x2 + y2) 2Chng minh 2): Ta c:xy(x2 + y2) (x

2 + y2 + 2xy)2

8= 2

206

2a2b2 3abtc (2t4 3t3c) = (t2 ab)(2t2 + 2ab 3tc) 0,t dn n

(a + 2b + 2c)2

2a2 + bc+

(b + 2c + 2a)2

2b2 + ca 2(4t

2 ab + 2tc)22a2b2 3abtc + 4t3c

2(3t2 + 2tc)2

2t4 3t3c + 4t3c =2(3t + 2c)2

2t2 + tc.

Mt khc, ta li c(c + 2c + 2b)2

2c2 + ab (4t + c)

2

t2 + 2c2. ()

Kt hp hai nh gi () v (), ta a bi ton v vic chng minh

2(3t + 2c)2

2t2

+ tc

+(4t + c)2

t2

+ 2c2

25.

Sau khi thu gn, ta c bt ng thc hin nhin ng

c(31t + 16c)(t c)2t(2t + c)(t2 + 2c2)

0.

Bi ton c chng minh xong. 21.22 Cho a,b,c,d l cc s thc khng m tha mn a2 + b2 + c2 + d2 = 1. Chng minh rng:

a

b2 + 1+

b

c2 + 1+

c

d2 + 1+

d

a2 + 1 4(a

a + b

b + c

c + d

d)2

5

Li gii. p dng bt ng thc Cauchy - Schwarz, ta c

ab2 + 1

+ bc2 + 1

+ cd2 + 1

+ da2 + 1

= a3

a2b2 + a2+ b

3

b2c2 + b2+ c

3

c2d2 + c2+ d

3

d2a2 + d2

(a

a + b

b + c

c + d

d)2

a2 + b2 + c2 + d2 + a2b2 + b2c2 + c2d2 + a2d2.

Nh vy, kt thc chng minh, ta cn ch ra rng

a2 + b2 + c2 + d2 + a2b2 + b2c2 + c2d2 + a2d2 54

,

hay (a2 + c2)(b2 + d2) 14

. Tuy nhin y li l nh gi ng v theo bt ng thc AM-GM:

(a2 + c2)(b2 + d2) (a2 + c2 + b2 + d2)2

4=

1

4,

do vy bt ng thc ban u c chng minh xong.

Bi ton kt thc. 2

1.23 Cho x,y,z l cc s thc thuc on [0, 1]. Chng minh rng:x

3

1 + y3+

y3

1 + z3+

z3

1 + x3 3

3

1 + xyz

Li gii. Do x,y,z [0, 1] nn ta c

x3

1 + y3+

y3

1 + z3+

z3

1 + x3 1

3

1 + y3+

13

1 + z3+

13

1 + x3.

27


28/116

rng theo bt ng thc Holder, ta c nh gi sau vi mi s thc dng a,b,c:(a + b + c)3 9(a3 + b3 + c3),

hay(a + b + c) 39(a3 + b3 + c3).

S dng nh gi ny, ta c1

3

1 + y3+

13

1 + z3+

13

1 + x3 3

9

1

1 + y3+

1

1 + x3+

1

1 + z3

.

Nh vy, kt thc chng minh, ta cn ch ra rng1

1 + y3

+1

1 + x3

+1

1 + z3

3

1 + xyz

. (

)

rng vi hai s thc a, b thay i trong on [0, 1] ta lun c1

1 + a2+

1

1 + b2 2

1 + ab=

(ab 1)(a b)2(1 + a2)(1 + b2)(1 + ab)

0.S dng nh gi ny, ta c

1

1 + x3+

1

1 + y3+

1

1 + z3+

1

1 + xyz 2

1 +

x3y3+

2

1 +

z4xy 4

1 + xyz.

Do vy nh gi () c chng minh, dn n bt ng thc ban u ng.


1.24 Cho a,b,c l cc s thc dng thay i bt k. Chng minh rng:a2

b + c+

b2

a + c+

c2

a + b a + b + c

2

Li gii 1. p dng bt ng thc Cauchy-Schwartz, ta ca2

b + c+

b2

a + c+

c2

a + b (a + b + c)

2

2(a + b + c)=

a + b + c

2.


Li gii 2. p dng bt ng thc AM-GM cho hai s dng, ta ca2

b + c+

b + c

4 a.

Cng v theo v nh gi ny vi hai nh gi tng t khc, ta c:a2

b + c+

b2

a + c+

c2

a + b+

a + b + c

2 a + b + c,

t ta thu c bt ng thc cn chng minh.

Bi ton kt thc. 2

Li gii 3. Bt ng thc ban u mang tnh i xng gia cc bin, do khng mt tnh tngqut, ta gi s a b c. Khi ta c

1

b + c 1

a + c 1

a + b.

Nh vy, theo bt ng thc Chebyshev, ta ca2

b + c+

b2

a + c+

c2

a + b 1

3.(a2 + b2 + c2).(

1

a + b+

1

b + c+

1

a + c).

n y ta p dng hai nh gi c bn

x2 + y2 + z2 (x + y + z)2

3v

28

9.37 Cho bai s thc dng x; y; zc tha mnxy + yz + zx = xyz . Chng minh rng:1

x2+

1

y2+

1

z2 3

x2y+

3

y2z+

3

z2xLi gii.

t a =1

x, b =

1

y, c =

1

zth a + b + c = 1

Bt ng thc cho tng ng vi:a2 + b2 + c2

3(a2b + b2c + c2a)

(a2 + b2 + c2)(a + b + c) 3(a2b + b2c + c2a) (do a + b + c = 1). (a3 + ab2) +( b3 + bc2) +( c3 + ca2) 2a2b +2 b2c +2 c2a(ng theo bt Cauchy cho 2 s dng).2

9.38 Cho hai s thc dng a; b; c . Chng minh rng:1

a + b+

1

b + c+

1

c + a+

1

2 3

abc (a + b + c +

3

abc)2

(a + b)(b + c)(c + a)

Li gii.

p dng bt ng thc cauchySchawrz, ta c:1

a + b+

1

b + c+

1

c + a+

1

2 3

abc= c2

c2(a + b)+

3

abc2

2abc

(a + b + c + 3

abc)2

(a + b)(b + c)(c + a).


9.39 Cho hai s thc a; b; c tha mn a2 + b2 + c2 =3

ab + bc + ac . Chng minh rng:a + b + c 3

Li gii.

D thy ab + bc + ca a2 + b2 + c2T ta c:

2

a2 + b2 + c2 3a2 + b2 + c2tng ng vi

(a2 + b2 + c2)3 (a2 + b2 + c2)2tng ng

a2 + b2 + c2 1tng ng

1 a2 + b2 + c2 (a + b + c)2

3

hay a + b + c 3. 2

9.40 Cho ba s thc dng a; b; c cng ln hn 2 v tho mn1

x+

1

y+

1

z= 1. Chng minh

rng:xyz + 9 4.(x + y + z)

Li gii.

Trong 3 s x,y,z phi c 2 s cng ln hn hoc cng nh hn so vi 3, gi s l x vy.Khi (x 3)(y 3) 0 xy + 9 3(x + y)Mt khc t gi thit suy ra xyz = xy + z + zx

205


29/116

Php chng minh hon tt. 29.35 Cho ba s thc dng a; b; c tha mna + b + c = 3. Chng minh rng: 1

1 + ab

9

2(

a +

b +

c)

Li gii.

Bt ng thc cho tng ng vi: ab1 + ab

3 92(

a +

b +

c)

p dng bt ng thc AM-GM, ta c:

ab1 + ab ab

2ab=

1

2

ab

By gi ta ch cn chng minh: ab +

9a +

b +

c 6

n gin ta t x =

a; y =

b; z =

c.Khi : x2 + y2 + z2 = 3 v:xy + yz + zx +

9

x + y + z 6

Tng ng xy + yz + zx +(x2 + y2 + z2)

3(x2 + y2 + z2)

x + y + z 2(x2 + y2 + z2)

(em thun nht bt ng thc). Do bt ng thc thun nht nn chun ha a + b + c = 1.t xy + yz + zx = A. Ta c:x2 + y2 + z2 = 1 2A, A 1

3 1 9A2 0ng thc xy ra khi v ch khi a = b = c = 1. 2

9.36 Cho hai s thc x; y tha mn xy = 0 v xy(x + y) = x2xy + y2. Tm gi tr ln nhtca biu thc:

A =1

x3+

1

y3

Li gii.

Vi ch :

A =1

x3+

1

y3=

x3 + y3

x3y3=

(x + y)(x2 xy + y2)x3y3

= 1

x+

1

y2

t a = 1x

, b =1y

.iu kin ca bi ton c th vit li nh sau :

a + b = a2 ab + b2 v ta cn tm gi tr nh nht ca biu thc :A = (a + b)2 vi nh gi:a2 ab + b2 = 1

4(a + b)2 +

3

4(a b)2 1

4(a + b)2

Ta thu c :a + b 14

(a + b)2hay l 0 a + b 4T suy ra A 16. Vy gi tr nh nht ca A l16 t c khi a = b hay x = y = 1

2. 2

204

1

x+

1

y+

1

z 9

x + y + z c

a2

b + c+

b2

a + c+

c2

a + b 1

3.(a + b + c)2

3.

9

2(a + b + c)=

a + b + c

2.


1.25 Cho a,b,c l cc s thc dng thay i bt k. Chng minh rng:4

3 + a4 + 4

3 + b4 + 4

3 + c4 4108(a + b + c)Li gii. p dng bt ng thc Holder, ta c

(1 + 3)(1 + 3)(1 + 3)(a4 + 3) (a + 3)4,t suy ra

4

3 + a4 3 + a4

64.

Thit lp cc nh gi tng t v cng li, ta c4

3 + a4 + 4

3 + b4 + 4

3 + c4 9 + a + b + c4

64.

Hn na, theo bt ng thc AM-GM, ta c9 + a + b + c = 3 + 3 + 3 + (a + b + c) 4 427(a + b + c),

nh vy4

3 + a4 + 4

3 + b4 + 4

3 + c4 44

27(a + b + c)4

64= 4

108(a + b + c).


1.26 Cho a, b l cc s thc dng tho mn ab 1. Chng minh rng:11 + a2

+1

1 + b2 2

1 + ab

Li gii. Thc hin php bin i tng ng, ta thu c dy cc nh gi sau:2 + a2 + b2

a2b2 + a2 + b2 + 1 2

1 + ab,

2 + 2ab + a3b + b3a + a2 + b2 2a2b2 2a2 2b2 2 0,(ab 1)(a b)2 0.

nh gi cui cng ng do ab 1, do vy bt ng thc ban u c chng minh.

Bi ton kt thc. 2

1.27 Cho a,b,c l cc s thc dng tho mn a + b + c = 1. Chng minh rng:c + ab

a + b

+a + bc

b + c

+b + ac

a + c 2

Li gii. rng ta c

c + ab = c(a + b + c) + ab = (c + a)(c + b),

do vy bt ng thc cn chng minh tng ng vi(c + a)(c + b)

a + b+

(b + a)(b + c)

a + c+

(a + b)(a + c)

b + c 2.

p dng nh gi c bn x2 + y2 + z2 xy + yz + zx, ta thy nh gi trn ng do(c + a)(c + b)

a + b+

(b + a)(b + c)

a + c+

(a + b)(a + c)

b + c b + c + a + b + c + a = 2.

29


30/116


1.28 Cho x,y,z l cc s thc dng tho mn 2x + 3y + z = 1 . Tm gi tr nh nht cabiu thc:

P = x

tonghop bdt cua mathscope

Documents

Transcript of tonghop bdt cua mathscope