Phuong Phap Don Va Giam Bien Trong Bat Dang Thuc

5/24/2018 Phuong Phap Don Va Giam Bien Trong Bat Dang Thuc

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I HC THI NGUYN

TRNG I HC KHOA HC

INH NGC QUANG

PHNG PHP DN V GIMBIN TRONG BT NG THC

LUN VN THC S

Chuyn ngnh : PHNG PHP TON S CPM s:60 46 40

Gio vin hng dn:GS.TSKH. NGUYN VN MU

THI NGUYN, 2012

S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn


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2

Mc lc

M u 4

1 Mt s dng bt ng thc c in v phng php gimbin 71.1 Cc bt ng thc hai bin lin quan n gi tr trung bnh . 71.2 Cc bt ng thcnbin lin quan n gi tr trung bnh . . 101.3 Phng php gim bin trong bt ng thc i s . . . . . . 12

1.3.1 Tam thc bc hai . . . . . . . . . . . . . . . . . . . . 131.3.2 Phng php tam thc bc hai nh hng . . . . . . 141.3.3 Gim bin trong bt ng thc i s . . . . . . . . . 15

2 gn u v phng php dn bin 212.1 gn u . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2 Hm li v biu din ca hm li. . . . . . . . . . . . . . . . 25

2.2.1 Hm li, lm. . . . . . . . . . . . . . . . . . . . . . . 252.2.2 Biu din hm li, lm . . . . . . . . . . . . . . . . . 27

2.3 Phng php dn bin . . . . . . . . . . . . . . . . . . . . . 302.3.1 Dn bin tng qut . . . . . . . . . . . . . . . . . . . 332.3.2 Mt s nh l v dn bin . . . . . . . . . . . . . . . 35

3 Mt s p dng 393.1 Mt s k thut thng dng trong gii bi ton bt ng thc 39

3.1.1 K thut chun ha . . . . . . . . . . . . . . . . . . . 393.1.2 K thut sp th t b s . . . . . . . . . . . . . . . 40

3.2 K thut dn bin. . . . . . . . . . . . . . . . . . . . . . . . 40

3.2.1 Dn cc bin bng nhau. . . . . . . . . . . . . . . . . 41



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3.2.2 Dn bin ra bin. . . . . . . . . . . . . . . . . . . . . 45

3.2.3 Dn bin trong lp hm li, lm . . . . . . . . . . . . 483.2.4 Dn bin trong lp hm n iu. . . . . . . . . . . . 50

Kt lun 52

Ti liu tham kho 53



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4

M u

Bt ng thc l mt vn kh c in ca ton hc s cp v ang

ngy cng pht trin, y cng l mt trong nhng phn ton hc s cpp v th v nht, v th lun cun ht rt nhiu ngi quan tm. Bt ngthc lun gi v tr quan trng trong cc k thi hc sinh gii, thi i hc,Olympic quc gia v quc t. im c bit v n tng nht ca bt ngthc trong ton hc s cp l c rt nhiu bi ton kh, thm ch l rtkh nhng lun c th gii c bng nhng kin thc c s, ch yu sdng cc php bin i, nh gi s cp thu c kt qu.

Ngy nay, c rt nhiu cc phng php chng minh bt ng thc thngdng nh: phng php s dng cc bt ng thc c in, phng phptam thc bc hai, phng php dng o hm, phng php phn tch bnhphng S.O.S., phng php vc t, phng php ta ,...Trong nhngnm gn y, GS.TSKH. Nguyn Vn Mu [1] gii thiu phng phptam thc bc hai nh hng. y l c s c phng php gim bin.Phng php gim bin c th pht biu bng li nh sau: Phng phpny da vo lt ct v php bin i ng dng gim s bin. Thng

thng, phng php ny hiu qu trong trng hp ba bin chuyn v biuthc dng hai bin. Cng trong khong thi gian ny, TS. Trn Nam Dngv Gabriel Dospinescu [3] gii thiu v trnh by phng php dn bin(Mixing variables). y l phng php rt quan trng v hiu qu trongvic chng minh cc bt ng thc phc tp. Phng php dn bin cth pht biu mt cch n gin nh sau: chng minh bt ng thcf(x1, x2, . . . , xn) 0chng ta i chng minh bt ng thc

f(x1, x2, . . . , xn) fx1+x22

,x1+x22

, x3, . . . , xn

.



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Sau chng ta i chng minh bt ng thc

f

x1+x2

2 ,

x1+x22

, x3, . . . , xn

0.

Bt ng thc sau ch cn n 1bin v n gin hn bt ng thc banu (cnbin).

Mc ch ca lun vn ny l trnh by li mt cch tng quan, c hthng cc kin thc c s v mt s bt ng thc c bn lin quan n gi

tr trung bnh, bt ng thc Cauchy lin quan n tam thc bc hai v xtn phng php gim bin, bt ng thc Karamata, gn u ca b sv xt nh l dn bin tng qut nh l h qu ca chng. Tip theo xtmt s ng dng ca phng php dn bin trong cc bi ton chng minhbt ng thc thng gp trong cc k thi hc sinh gii v k thi Olympic.

Lun vn gm phn m u, kt lun, ti liu tham kho v 3chng.

Chng 1, trnh by mt s bt ng thc c bn lin quan n gi tr trung

bnh v bt ng thc Cauchy lin quan n tam thc bc hai. Cc kinthc ny l c s trnh by cc ni dung quan trng cui chng 1v trong chng 2.

Chng 2, trnh by v gn u, mt s khi nim v tnh cht quantrng ca hm li, lm, t i n trnh by phng php dn bintng qut. Phng php dn bin v c bn l cch thc lm gim bintrong bt ng thc i s.

Chng 3, trnh by mt s p dng ca phng php dn bin v gim bingii cc bi ton bt ng thc 3 bin, 4 bin.

Qua y, chng ti xin c gi li cm n su sc n thy gio, ngihng dn khoa hc ca chng ti, GS.TSKH. Nguyn Vn Mu, ngi a ra ti v tn tnh hng dn trong sut qu trnh nghin cu cachng ti. ng thi chng ti cng chn thnh cm n cc thy c trongkhoa Ton - Tin hc trng i hc Khoa hc, i hc Thi Nguyn, to mi iu kin v ti liu v th tc hnh chnh chng ti hon thnh



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lun vn ny. Chng ti cng gi li cm n n bn b, c bit l cc bn

hc vin trong lp Cao hc Ton K4, ng vin gip chng ti trongqu trnh hc tp v lm lun vn.

Do thi gian hn hp v khi lng kin thc ln, chc chn lun vnkhng th trnh khi nhng thiu st. Chng ti rt mong c s ch botn tnh ca cc thy c v bn b ng nghip. Chng ti xin chn thnhcm n.

Thi Nguyn, nm 2012Hc vin

inh Ngc Quang



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Chng 1

Mt s dng bt ng thc c in

v phng php gim binTrong chng ny chng ti trnh by mt s bt ng thc c bn lin

quan n gi tr trung bnh v bt ng thc Cauchy lin quan n tam thcbc hai phc v cho vic trnh by cc ni dung chnh ca lun vn trongcc phn sau.

Cc vn c trnh by trong chng ny c tham kho v trch dnch yu ti mt s ti liu [1], [3].

1.1 Cc bt ng thc hai bin lin quan n gi trtrung bnh

Xut pht t bt ng thc c bnx2 0, x. Khi

(x

y)2

0

x2 +y2 + 2xy

4xy

(x+y)2

4xy,

x,y.

Suy rax+y

2 xy, x, y 0. (1.1)

Bt ng thc (1.1) l mt bt ng thc quen thuc chng trnh tonph thng. Ta gi l bt ng thc gia trung bnh cng v trung bnh nhn(gi ngn gn l bt ng thc AM-GM1) vi 2 binx, y.

1Arithmetic Mean - Trung bnh cng, Geometric Mean - Trung bnh nhn.



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nh l 1.1 (Bt ng thc AM-GM vi 2 bin).Vi x1, x2 khng m, tac

x1x2x1+x22

. (1.2)

Du ng thc xy ra khi v ch khi x1= x2.Chng minh. Vix1, x2khng m, bnh phng 2 v bt ng thc (1.2)c

4x1x2 (x1+x2)2 4x1x2 x21+x22+ 2x1x2 0 (x1 x2)2

ng vi mix1, x2.

T bt ng thc (1.1) ta thc hin mt vi bin i

xyx+y

2 2xy

x+y xy

21

x+

1

y

xy, x, y 0. (1.3)

Bt ng thc (1.3) l mt H qu trc tip ca bt ng thc AM-GMvi 2 bin, v c gi l bt ng thc gia trung bnh nhn v trung bnhiu ha (gi ngn gn l bt ng thc GM-HM2) vi 2 binx, ykhng m.

H qu 1.1(Bt ng thc GM-HM vi 2 bin).Chox1, x2l cc s thckhng m, ta c

2

1x1

+ 1x2

x1x2. (1.4)

Du ng thc xy ra khi v ch khi x1= x2.

Chng minh. S dng bt ng thc (1.2) i vi x:= 1

x, y:=

1

y, ta c

1

x.1

y

1x

+ 1y

2 ,

2

Harmonic - Trung bnh iu ha.



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hay2

1x

+ 1y

xy.

Ta c iu cn chng minh.

T bt ng thc (1.1), bnh phng 2 v ta c

xy(x+y)2

4 ,

hay

2xy x2 +y2 (x+y)2 x2 y2

2 x

2 +y2

2 (x+y)

2

4 x

2 +y2

2 ,

ly cn bc hai 2 v ta c

21

x+

1

y

xy, x, y 0. (1.5)

Bt ng thc (1.3) l mt H qu ca bt ng thc AM-GM vi 2 bin,v c gi l bt ng thc gia trung cng v trung bnh bc hai (gi ngngn l bt ng thc AM-QM3) vi 2 bin x, ykhng m.

H qu 1.2 (Bt ng thc AM-QM vi 2 bin). Cho x, y l cc s thckhng m, ta c

x+y

2 x2 +y2

2

. (1.6)

Du ng thc xy ra khi v ch khi x= y.Chng minh. Vi x, y khng m, bnh phng 2v bt ng thc (1.6)ta c

(x+y)2

4 x

2 +y2

2 0 x

2 2xy+y24

0 (x y)2

ng vi mix, y.3Quadratic mean - Trung bnh bc hai (ton phng).



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T nhng chng minh trn chng ta rt ra c chui bt ng thc vi

2 binx, ykhng m nh sau

min{x, y} 21

x+

1

y

xyx+y2

x2 +y2

2 max{x, y}. (1.7)

1.2 Cc bt ng thc n bin lin quan n gi tr

trung bnhnh l 1.2(Bt ng thc AM-GM (Theo [3])).Cho x1, x2, . . . , xnl ccs thc khng m, n 1, khi

n

x1x2 . . . xnx1+x2+ +xn

n . (1.8)

Du ng thc xy ra khi v ch khi x1= x2= =xn.Chng minh. (Phng php chng minh quy np Cauchy)4

Vin= 1, bt ng thc (1.8) hin nhin ng.Vin = 2, ta c bt ng thc (1.2) chng minh l ng nh l1.1.Gi thit quy np: Gi s bt ng thc (1.8) ng vi ns thc khngmx1, x2, . . . , xn, n 1.Cho 2ns thc khng m x1, x2, . . . , xn, xn+1, . . . , x2n, ta xt

x1+x2+ +x2n2n

=1

2

x1+x2+ +xn

n +

xn+1+xn+2+ +x2nn

nx1x2 . . . xn+ nxn+1xn+2 . . . x2n

2 ( nx1x2 . . . xn nxn+1xn+2 . . . x2n)

1

2 ,

hayx1+x2+ +x2n

2n 2nx1x2 . . . x2n. (1.9)

T trng hpn= 1,n= 2v (1.9) suy ra (1.8) ng vi n= 2k, k 1.y chnh l quy np theo hng ln trn.

4y l kiu quy np theo cp hng (ln-xung) do Cauchy xut nm 1821 (Cauchy A.L., cours

dAnalyse de lEcole Royale Polytechnique,Ire partie, Analyse algebrique, Paris, Debure, 1821) chngminh nh l AM-GM[3].



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(Quy np hng xung di) Gi s bt ng thc (1.8) ng vins thc

khng m. Ta chng minh n ng vi n 1s khng mx1, x2, . . . , xn1.Xt

x1+x2+ +xnn

,

thay binxnbi x1+x2+ +xn1

n 1 ta c

x1+x2+ +xn1+ x1+x2++xn1n1n

x1x2 . . . xn1x1+x2+ +xn1

n

1

1

n

,

hay

x1+x2+ +xn1n 1

x1x2 . . . xn1

x1+x2+ +xn1

n 1 1

n

.

Nng ly tha bc nc 2 v ta cx1+x2+ +xn1

n 1n

x1x2 . . . xn1

x1+x2+ +xn1n 1

.

Suy ra

x1+x2+ +xn1n 1

n1 x1x2 . . . xn1.

Ly cn bc n 1c 2 v ta cx1+x2+ +xn1

n 1 n1

x1x2 . . . xn1.

Ta c iu cn chng minh.

H qu trc tip ca bt ng thc AM-GM l bt ng thc GM-HM.

H qu 1.3 (Bt ng thc GM-HM). Cho x1, x2, . . . , xn l cc s thckhng m,n 1, khi

n1

x1+

1

x2+ + 1

xn

nx1x2 . . . xn. (1.10)

Du ng thc xy ra khi v ch khi x1= x2= =xn.Chng minh. S dng bt ng thc AM-GM i vi b sxk :=

1

xk, (k=

1, 2, . . . , n), ta c ngay bt ng thc GM-HM.



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H qu ca bt ng thc AM-GM l bt ng thc AM-QM. Tng

t nh chng minh bt ng thc AM-GM, bng phng php chng minhquay np Cauchy ta c th chng minh bt ng thc AM-QM.

H qu 1.4 (Bt ng thc AM-QM). Cho x1, x2, . . . , xn l cc s thckhng m,n 1, khi

x1+x2+ +xnn

n

x21+x22+ +x2n

n . (1.11)

Du ng thc xy ra khi v ch khi x1= x2= =xn.T nhng bt ng thc (1.8), (1.10), (1.11) v t chui bt ng thc

(1.7), chng ta tip tc rt ra c chui bt ng thc vinbinx1, x2, . . . , xnkhng m nh sau

min{x1, x2, . . . , xn} n1x1

+ 1

x2+ + 1

xn

nx1x2 . . . xn

x1+x2+ +xnn

n

x21+x22+ +x2n

n max{x1, x2, . . . , xn}.

(1.12)

1.3 Phng php gim bin trong bt ng thc is

y, chng ti trnh by mt phng php lm gim bin trong btng thc i s. Phng php ny da vo lt ct v php bin i ngdng gim s bin. Thng thng, phng php ny hiu qu trong trnghp ba bin chuyn v biu thc dng hai bin.

Trong mc ny chng ti s trch dn mt s kin thc v tam thc bchai v cc nh l v du ca tam thc bc hai, t trnh by v phngphp gim bin trong bt ng thc i s.

Cc vn trnh by trong mc ny c chng ti tham kho v trchdn ch yu t ti liu [1].



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1.3.1 Tam thc bc hai

Chng ta vn xut pht t bt ng thc c bn

x2 0, x R. (1.13)Du ng thc xy ra khi v ch khi

x= 0.

Gn vi bt ng thc (1.13) l bt ng thc dng sau:

(x1 x2)2 0, x1, x2 R,hay

x21+x22 2x1x2, x1, x2 R.

Du ng thc xy ra khi v ch khi x1= x2.Xt tam thc bc hai

f(x) =ax2 +bx +c, a = 0.

Khi af(x) =

ax+

b

2

2

4,

vi =b2 4ac.T y, chng ti trch dn mt s kt qu sau.

nh l 1.3(Theo [1]).Xt tam thc bc haif(x) =ax2 + bx + c, a = 0.i) Nu< 0th af(x)> 0,

x

R.

ii) Nu = 0th af(x) 0, x R.Du ng thc xy ra khi v chkhix= b

2a.

iii) Nu> 0 th af(x) =a2(x x1)(x x2)vi

x1,2= b2a

2|a| . (1.14)

Trong trng hp ny, af(x) < 0khi x

(x1, x

2) v af(x) > 0 khi

x < x1hocx > x2.



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nh l 1.4 (nh l o (Theo [1])).iu kin cn v tn ti s sao choaf()< 0l > 0vx1< < x2, trong x1,2l cc nghim caf(x)xc nh theo (1.14).

nh l 1.5(Theo [1]).Vi mi tam thc bc haif(x)c nghim thc utn ti mt nguyn hm F(x)l a thc bc ba, c ba nghim u thc.

nh l 1.6(Theo[1]).Tam thc bc haif(x) = 3x2 + 2bx+cc nghim(thc) khi v ch khi h s b, cc dng:

b= ++,c= ++, (1.15)

trong , , R.

1.3.2 Phng php tam thc bc hai nh hng

Xt a thc thun nht bc hai hai bin (xem nh tam thc bc hai ivix)

F(x, y) =ax2 +bxy+cy2, a = 0, (1.16) := (b2 4ac)y2.

Khi , nu 0thaF(x, y) 0, x, y R.

Vy khib2 4acva


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gi tr ca n. Khi tm gi tr ln nht (nh nht) ca biu thc

cho, ta ch cn quan tm n cc gi tr ln hn (nh thua) ca biu thcm thi.

p dng vo cc bi ton cc tr ca mt s dng ton bc hai, ta sdng tnh cht ca dng phn thc bc hai

y=a1x

2 +b1x+c1a2x2 +b2x+c2

,

vi iu kina2>0, f2(x) =a2x

2 +b2x+c2>0, x R.Nhn xt 1.1. phng php nu trn, a thc thun nht bc

hai hai bin (1.16) c xem nh tam thc bc hai i vi x. Khi ,t bi ton hai bin x, y, binyc coi nh tham bin cho trc, thta ch phi lm vic vi bin x. Vy mc tiu chnh ca phng phpny l lm gim s bin a bi ton v dng tam thc bc hai, v

c th gii quyt c vi cc kin thc v du ca tam thc bc hai. c trng ca phng php ny l gii quyt cc bi ton tm cc tr

v gn lin vi cc kin thc v du ca tam thc bc hai.

Ta c th trnh by tng minh phng php ny thng qua bi ton sau

1.3.3 Gim bin trong bt ng thc i s

Bi ton 1.1(Theo [1]).Tm gi tr ln nht v gi tr nh nht ca hmsy=

a1x2 +b1x+c1

a2x2 +b2x+c2,

vi iu kin

a2>0, f2(x) =a2x2 +b2x+c2>0, x R.

Gii. Nhn xt rng khix = 0th y(0) = c1c2

v khi x th y a1a2

.

Tip theo, ta xt cc gi tr y= c1c2 vy= a1

a2.



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Gi syl mt gi tr ca biu thc, y=

c1

c2v y

=

a1

a2. Khi phng

trnh tng nga1x

2 +b1x+c1a2x2 +b2x+c2

=y

phi c nghim, hay phng trnh

(a2y a1)x2 + (b2y b1)x+ (c2y c1) = 0 (1.17)

phi c nghim.Do (1.17) l phng trnh bc hai nn iu ny tng ng vi

= (b2y b1)2 4(a2y a1)(c2y c1) 0,

hay

g(y) := (b22 4a2c2)y2 + 2(b1b2+ 2a2c1+ 2a1c2)y+b21 4a1c1phi c nghim. V g(y)c b22

4a2c2 < 0nn theo nh l o ca tam

thc bc hai1.4, th

= (b1b2+ 2a1c2+a2c1)2 (4a1c1 b21)(4a2c2 b22) 0 (1.18)

vy1 y y2,

vi

y1,2=

b1b2+ 2a2c1+ 2a1c2

b22 4a2c2 ,vc tnh theo cng thc (1.18).

Suy ra max y = y2v min y =y1, t c khi ng vi mi j(j = 1, 2),xy ra ng thi

= (b2yj b1)2 4(a2yj a1)(c2yj c1) = 0,xj = 1

2

b2yj b12a2yj a1 .

Sau y, chng ti trch dn v d minh ha sau.



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V d 1.1(Theo [1]).Chox, yl cc s thc sao cho

2x2 +y2 +xy 1.

Tm gi tr nh nht ca biu thc

M=x2 +y2.

Gii. t2x2 +y2 +xy=a, a 1. Khi

Ma

= x2

+y2

2x2 +y2 +xy.

Nu y= 0th Ma

=1

2.

Nu y= 0, tt= xy

, suy ra

M

a =

t2 + 1

2t2 +t+ 1.

Ta ch cn xc nh cc gi trM

a 1

2

2

(v ta ang cn

chng minhM 12

).

tx

z =,

y

z =. Khi

(1.19) (2 + 22 + 5)z2 = 1 (1.20)

M=z2(++) = ++

2 + 22 + 5, M >

1

2

2.

M.2

(+ 1)+M(22

+ 5) = 0, (coil n) (1.21)phi c nghim. Suy ra

= (+ 1)2 4M(2M2 + 5M) 0

(1 8M2)2 + 2(2M+ 1)+ 1 20M2 0. (1.22)Do

= (2M+ 1)2

(1

8M2)(1

20M2)

0

4M+ 32M2 160M4 0 4M(1 + 8M 40M3) 0

12

2< M 1

2.

Bi ton c chng minh.

Bi ton1.2c th trnh by yu cu ca bi ton theo cch khc.



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Bi ton 1.3.Chox2 + 2y2 + 5z2 = 1. (1.23)

Tm gi tr ln nht ca

M=xy +yz+zx.

Gii. Gii tng t li gii ca Bi ton1.2. Cui cng, khi ta tm ra M

t gi tr ln nht l M=1

2, th ta tnh ,, z, x, ybng cch th M=

1

2

ln lt nh sau.ThMvo (1.22) ta tnh ra.ThM, vo (1.21) ta tnh ra.ThM, , vo (1.20) ta tnh raz.T , suy ra x, y.



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Chng 2

gn u v phng php dn bin

Trong chng ny, chng ti trnh by v gn u, hm li v biu dinca hm li, t trnh by phng php dn bin dng tng qut.

Cc vn trnh by trong chng ny c chng ti tham kho v trchdn ch yu t mt s ti liu[1], [3], [8], [4].

2.1 gn u

Xut pht t vic nhn thy cc gi tr trung bnh u bng nhau khi vch khi cc s hng trong dy s thc khng m bng nhau hay cc cp strong dy s thc khng m bng nhau. iu ny lm chng ti ngh ti victm hiu v cp gi tr gn bng nhau trong dy s v s sp xp th t cacc cp gi tr.

T bt ng thc (1.2)

x21+x

22 2x1x2, x1, x2 R.

Ta suy ra vi mi cp s khng m x, yvi tng bng 1 cho trc th tch xy

t gi tr ln nht bng1

4khix = y =

1

2. Tuy nhin khi x, ybin i trong

mt min v trong min v xkhcy th chng ch t c v th v trxvygn nhau nht. T y ta pht biu khi nim gn u nh sau.

nh ngha 2.1(Theo [1]).Xt cc cp s khng m x, y. Ta gi hiu

(x, y) := max{x, y} min{x, y}



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l lch ca cp sx, yhay l gn u ca cp s x, y.

Nu(x, y) = 0th x= y, ta gix, yl cp u.

Nux =y th(x, y)> 0, ta gi gn u cax, ybng(x, y). phn chia cc trng hp r rng hn, chng ta pht biu nh ngha

2.1cho cc trng hp ring nh sau.Nhn xt rng, i vi cc cp s dng c chung tng th ta c th pht

biu th t cc cp rng tch xy t gi tr ln nht trong trng hpcp s l u, tc l x= y.

nh ngha 2.2. i) Xt cc cp s khng m x, y vi tng khng i( n gin, ta chn x+y = 1). Ta gi hiu



ii) Cpx1, y1c gi l gn u hn ( lch nh hn) cp x2, y2(haycpx2, y2c gi l xa u hn cp x1, y1) nu

(x1, y1) (x2, y2).

Nh vy theo nh ngha2.2th cp x, ylun c lch (x, y)0v lch bng 0 khi v ch khi x= y , khi cp x, yl cp gn u nht vcc cp0, 1v1, 0s l nhng cp xa u nht.

Nhn xt rng, i vi cc cp s dng c chung tch th ta c th phtbiu th t cc cp rng tngx + yt gi tr nh nht trong trng hpcp s l u, tc l x= y.

nh ngha 2.3. i) Xt cc cp s khng mx, yvi tch khng i (n gin, ta chnxy=a >0). Ta gi hiu





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ii) Cpx1, y1c gi l gn u hn ( lch nh hn) cp x2, y2(hay

cpx2, y2c gi l xa u hn cp x1, y1) nu

(x1, y1) (x2, y2).Nh vy theo nh ngha2.3th cp x, ylun c lch (x, y)0v

lch bng 0 khi v ch khi x= y , khi cp x, yl cp gn u nht vcc cp0, 1v1, 0s l nhng cp xa u nht.

Th t sp c ngn ng "gn u dn" cho ta mt cch tip cn t

nhin vi nhiu bi ton ca thc tin. Sau y, chng ti trnh by mt biton lm v d.

Bi ton 2.1. Khi ta c mt cp s x, ydng c tng bng 7. Ta c mtlot cc b s: (1, 6), (2, 5), (3, 4)c tng bng 7. Tt c c chung mt ctrng

(x.y)

7

2

2.

Nu xem xt k ta thy cc tch

1.6< 2.5< 3.4.

Vy cp s c tng khng i th khixvy cng gn u nhau th tch cngln. Lc ny ta c th t sp cht ca b s .

Nhn xt 2.1.Khi ta c mt cp s t nhin x, yc tng l mt s l thcp s s khng bao gi l cp s nguyn bng nhau c. Khi , khinim gn u nht (m khng phi l u) s c ngha thc tin. T y

tin gn n trng hp u.T cc nh ngha2.1,2.2,2.3,ta suy ra c thm mt s tnh cht v

gn u ca cc cp s khng m.Vi quy c ca nh ngha2.2, ta c th sp th t cc tch xy vi tng

x+y = 1khng i theo ngn ng "gn u" nh sau.

nh l 2.1 (Theo [1]). Xt cc cp s khng m xj, yj, vi j = 1, 2vitng khng i ( n gin ta chnxj+yj = 1). Khi

x1y1 x2y2



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khi v ch khi cp x1, y1gn u hn cp x2, y2.Chng minh. Xt cc cp s khng m x, y vi tng bng 1 khng i.Khng mt tnh tng qut, coi 0 x < y. Vi mi > 0 nh, mbox + < y + (ch cn chn

0,

y x2

). D thy rng cpx + , y

gn u hn cp x, y cho. Ta ch cn chng minh rng

xy (x+)(y ) (2.1)

l .iu ny l hin nhin v (2.1) tng ng vi bt ng thc(yx) 0.

Vi quy c ca nh ngha2.3, ta c th sp th t cc tng x+y vitchxykhng i theo ngn ng "gn u" nh sau.

nh l 2.2(Theo [1]).Xt cc cp s khng mxj, yj, vij = 1, 2vi tchkhng i ( n gin ta chnxjyj = 1). Khi

x1+y1 x2+y2khi v ch khi cp x1, y1gn u hn cp x2, y2.Chng minh. Xt cc cp s dngx, yvi tch bng 1 khng i. Khngmt tnh tng qut, coi 0< x < y. Vi mi >1, m bo x


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Bi ton 2.2.Xt hai cp s (z, 2

z)v(y, 2

y)vi

1 x y 2,hoc

0 y z 1.Khi , theo nh ngha2.1, r rng cp s(z, 2 z)gn u hn so vi cps(y, 2 y).

T chng ta c mt nh l ng vai tr quan trng i vi hai cp

s sp c th t.nh l 2.3(H. W. Melaughlin, F. T. Metcalf (Theo [1])).Vi mi cp dys dng a = (a1, a2, . . . , an)sao cho 0 y z 1v b = (b1, b2, . . . , bn)sao cho1 z y 2hoc ta u c

nk=1

aykb2yk

nk=1

a2yk byk

nk=1

azkb2zk

nk=1

a2zk bzk

.

y chnh l mt dng ni suy bt ng thc Cauchy trong [0, 1].

2.2 Hm li v biu din ca hm li

2.2.1 Hm li, lm

nh ngha 2.4(Hm li).Mt hm s f : [a, b] Rc gi l li nuf(tx+ (1 t)y) tf(x) + (1 t)f(y), x, y [a, b], t [0, 1].

T nh ngha ta rt ra c mt s nhn xt quan trng nh sau.

Nhn xt 2.2. 1. Khix1 < x2thx= tx1+ (1 t)x2vi mi t [0, 1].u thuc (x1, x2)v

t= x x1x2 x1 , 1 t=

x2 xx2 x1 .

2. Nufkh vi hai ln th hm fli khi v ch khif(x) 0, x [a, b].3. Nu f li th flin tc. Ngc li, nu flin tc th tnh li ca f l

tng ng vi fx+y

2 f(x) +f(y)2 .



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Tnh li (Theo [8]) Mt hm sf(x)l lm ln (xung) trn on[a, b]R

nuf(x)nm di (trn) ng thng ni cc im(a1, f(a1))v(b1, f(b1)),

a a1 x b1 b.

Hm s lm ln v xung c gi tng ng l hm li v hm lm.Nu f l hm li trn on [a, b]v 1, 2, . . . nl nhng s khng m

c tng bng 1, th

1f(x1) +2f(x2) + +nf(xn) f(1x1+2x2+ +nxn), (2.3)vi bt k x1, x2, . . . , xntrong on [a, b].

Nu hm s l lm, bt ng thc s i chiu.Tng t, ta cng c nh ngha hm li (lm) trn cc tp(a, b), [a, b), (a, b].

Ta s s dng k hiu I(a, b) l nhm ngm nh mt trong bn tp hp[a, b], (a, b), [a, b), (a, b].

Tip theo, chng ti s trch dn mt s tnh cht ca hm li trnI(a, b).

Tnh cht sau cho php ta d dng kim chng tnh li (lm) i vi mthm s cho trc.

nh l 2.4 (Jensen (Theo [1])). Gi s f(x) lin tc trn [a, b]. Khi iu kin cn v hm s f(x)li trnI(a, b)l

f

x1+x2

2

f(x1) +f(x2)

2 , x1, x2 I(a, b). (2.4)

Chng minh. Nu f(x)l hm li trn I(a, b)th ta c ngay (2.4) bngcch chnt=1

2t nh ngha hm li.

Gi s c (2.4). Ta cn chng minh rng vit [0, 1]ta u c

f(tx1+ (1 t)x2) tf(x1) + (1 t)f(x2).

Nut Qta c th vit

t=m

q

, 1

t=

n

q

,



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trong m, nZ, q

Nv m+n = q. Bng phng php quy np ta c

ngay

f(tx1+ (1 t)x2) =f

mx1+nx2q

mf(x1) +nf(x2)q

=tf(x1) + (1 t)f(x2).

Nutl s v t th ta chn dy s hu t dng untrong khong (0, 1)cgii hn bng t

limn un =t.Khi , hin nhin dyvn:= 1 uncng nm trong(0, 1)v

limn vn= 1 t.

Theo chng minh trn ng vi trng hp thu t, th

f(unx1+vnx2) unf(x1) +vnf(x2), n N, x1, x2 I(a, b).Chuyn qua gii hn v s dng tnh lin tc ca f(x), ta thu c

f(tx1+ (1 t)x2) tf(x1) + (1 t)f(x2).nh l c chng minh.

2.2.2 Biu din hm li, lm

Trong phn ny, chng ti trch dn ch yu cc kt qu t [1]. n gin cch trnh by, ta ch xt lp cc hm li lm kh vi bc hai.

Nh vy, hmf(x)n iu tng trong I(a, b)khi v ch khi

f(x) 0, x I(a, b),v hmf(x)li trnI(a, b)khi v ch khi

f(x) 0, x I(a, b).T , ta c nhn xt, khi hm f(x)li trn I(a, b)th o hm bc nhtca n l mt hm n iu tng.

Do vy, ta c th pht biu tnh cht biu din hm li nh sau.



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nh l 2.5(Theo [1]).Hm f(x)li trnI(a, b)khi v ch khi tn ti hmg(x)n iu tng trong I(a, b)v s c (a, b), sao cho

f(x) =f(c) +

xc

g(t)dt.

y, chng ta c bit quan tm n mt dng biu din hm li, hmlm thng qua cc hm s bc nht, v lp hm ny n gin v d tnh ton

trn tp gi tr ca chng. rng, nu f(x)l hm li lin tc trn [a, b]v vi s t [0, 1]xyra ng thc

tf(a) + (1 t)f(b) =f(ta+ (1 t)b)thf(x)l hm s bc nht.

V vy, khi hm s f(x)li v kh vi trn I(a, b)th th ca n lunlun thuc na mt phng trn to nn bi tip tuyn ti mi im ty cho trc ca th . Ni cch khc, nu f(x)li trnI(a, b)th vi micpx0, x I(a, b), ta u c

f(x) f(x0) +f(x0)(x x0). (2.5)

Tht vy, ta c (2.5) tng ng vi

f(x0) f(x) f(x0)x x0 , khix > x0; x0, x I(a, b), (2.6)

vf(x0) f(x) f(x0)

x x0 , khix < x0; x0, x I(a, b). (2.7)

Cc bt ng thc (2.6), (2.7) l hin nhin (theo nh l Lagrange).D nhn thy rng (2.5) xy ra ng thc khi x0=x. Vy ta c th vit

(2.5) di dngf(x) = max

uI(a,b)[f(u) +f(u)(x u)].

Tng t, ta cng c biu din i vi hm lm.



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Khi hm s f(x)lm v kh vi trn I(a, b)th th ca n thuc na

mt phng di to bi tip tuyn ti mi im ty thuc th, tc lvi mi cp x0, x I(a, b), ta u c

f(x) f(x0) +f(x0)(x x0). (2.8)D nhn thy rng (2.8) xy ra ng thc khi x0 = x. Vy ta c th vit(2.8) di dng

f(x) = minuI(a,b)

[f(u) +f(u)(x u)]. (2.9)

Vy chng ta c mt dng biu din hm li v lm thng qua cc trca cc hm s bc nht ph thuc tham bin. Php biu din ny c gil biu din tuyn tnh (theo Bellman) v n ng vai tr quan trng nhl mt cng c hu hiu trong nhiu bi ton cc tr v ti u.

Tng t, ta cng c biu din i vi hm li nhiu bin. Gi sF(x1, x2,...,xn)l hm s thc li kh vi theo tng bin trn R. Khi , ng vi mi b s(z1, z2, . . . , z n)m

z1 z2 zn,ta u c

F(x1, x2, . . . , xn) F(z1, z2, . . . , z n) +n

i=1

(xi zi)Fzi

.

T suy ra

F(x1, x2, . . . , xn) = maxR(z) F(z1, z2, . . . , z n) +

n

i=1

(xi zi)Fzi .

Nhn vo nh ngha, cc nh l v hm li v biu din hm li, ta cth thy mc tiu a hm hai hay nhiu bin gi s nh

f(x1, x2, . . . , xn)

v hm mt bin gi s nh

nfx1+x2+ +xn

n ,hay ta gi ldn bin.



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2.3 Phng php dn bin

Phng php dn bin c t tng lm gim s bin trong bi ton, davo php nn gn u, n gip a bi ton v dng n gin hn.

Trc khi trnh by phng php dn bin tng qut, chng ti trch dnmt s kt qu c s.

nh l 2.6 (iu kin Schur (Theo [1])). iu kin cn v hai bdy s n iu gim{xk, yk :k = 1, 2, . . . , n}, tha mn cc iu kin

x1 y1x1+x2 y1+y2. . . . . . . . .x1+x2+ +xn1 y1+y2+ +yn1x1+x2+ +xn =y1+y2+ +yn

(2.10)

l gia chng c mt php bin i tuyn tnh dng

yi=n

j=1

aijxj , i= 1, 2, . . . , n ,

trong akl 0,

nj=1

akj = 1,n

j=1

ajl = 1; k, l= 1, 2, . . . , n .

Ta gi hai b dy s{xk, yk :k= 1, 2, . . . , n}tha mn iu kin (2.10)l dy {x1, x2,...,xn} tri hn dy {y1, y2,...,yn}, hay dy {y1, y2,...,yn} gnu hn dy{x1, x2,...,xn}.nh l 2.7(Karamata (Theo [1])).Cho hmy =f(x)c o hm cp hai

ti mix (a, b)sao cho f(x) 0, x [a, b]vf(x)> 0, x (a, b).Gi sx1, x2, . . . , xn v y1, y2, . . . , yn l cc s [a, b]ng thi tha mncc iu kin

x1 x2 xnvy1 y2 ynv

x1 y1x1+x2 y1+y2. . . . . . . . .

x1+x2+ +xn1 y1+y2+ +yn1x1+x2+ +xn y1+y2+ +yn



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Khi , ta lun c

nk=1

f(xk) = maxt1,...tnI(a,b)

ni=1

f(ti) +n

i=1

(xi ti)f(ti)

n

k=1

f(yk). (2.11)

Chng minh. S dng biu din i vi hm li ta c

f(x1) +f(x2) + +f(xn) =

= maxt1,...tnI(a,b) n

i=1f(ti) +

ni=1

(xi ti)f(ti) . (2.12)Khng mt tnh tng qut, ta gi thit b s t1, t2, . . . , tnI(a, b)cng lmt b s gim, tc l

t1 t2 tn.Khi , ta ch cn chng minh rng

x1f(t1) +x2f(t2) +

+xnf

(tn)

y1f(t1) +y2f(t2) + +ynf(tn).S dng bin i Abel

x1f(t1) +x2f(t2) + +xnf(tn) =

=S1[f(t1) f(t2)] +S2[f(t2) f(t3)] + +

+Sn1[f(tn1)

f(tn)],

viSk(x) :=x1+x2+ +xk.

V f(x) > 0 nn f(xk) f(xk1). Mt khc, do Sk(x) Sk(y)(k =1, 2, . . . , n 1)v v Sn(x) Sn(y)v f(tn) 0. Ta c ngay iu cnchng minh.

T y ta c c nh hng v dn bin trong lp hm li, lm, ctrnh by nh sau.



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Dn bin trong lp hm li, lm

Cho hai b s x1, x2, . . . , xnvy1, y2, . . . , yntha mn

x1 x2 xnvy1 y2 ynv

x1 y1x1+x2 y1+y2. . . . . . . . .x1+x2+

+xn

1

y1+y2+

+yn

1

x1+x2+ +xn y1+y2+ +yn.Khi , vi hm s f(x)tha mn

f(x) 0, x (a, b)

vf(x1) +f(x2) + +f(xn) f(y1) +f(y2) + +f(yn).

Ta lun c

f(x1) + f(x2) + + f(xn) 2f

x1+x22

+ f(x3) + + f(xn). (2.13)

Nhn vo (2.13), ta c th thy mc tiu ca chng ta l a hai hay nhiuhm mt bin

f(x1), f(x2), . . . , f (xn)

v mt hm mt bin gi s nh

nf

x1+x2+ +xnn

.

y l mt cng c dn bin tt. Chng ta c th s dng a tt ccc bin bng nhau (gi l dn bin ton cc). Tuy nhin, nhng bi tonnh vy khng phi l a s, v nhng bt ng thc s dng c dn bindng (2.13) th cc bin phi nm trong cc biu thc c lp nhau, cth vit thnh dng

f(x1) +f(x2) + +f(xn). (2.14)



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Nh trnh by trn, khng c nhiu bt ng thc c th vit thnh

dng (2.14), m ta s phi lm vic vi dng tng qut hn l

f(x1, x2, . . . , , xn).

Nh vy t tng chnh ca dn bin l thay v mong mun c ngay mtcch dn bin ton cc, chng ta hy vng c th tng bc n gin bi tonbng cch gim dn s bin.

2.3.1 Dn bin tng qut

Nhn xt rng, iu kin (2.10) cho ta mt thut ton nn u mt bs cho trc bng phng php bin i tuyn tnh. Ta nhn thy php biudin hm li (lm) mt bin c th m rng cho hm li nhiu bin. C thl, ng vi mi hm li F(x1, x2, . . . , xn), ta u c

F(x1, x2, . . . , xn) = max(t1,...,tn)

F(t1, t2, . . . , tn) +

ni=1

(xi ti)Fti

. (2.15)

S dng cch chng minh tng t chng minh nh l2.7, ta thu cnh l sau.

nh l 2.8(Theo [1]).Gi sF(x1, x2, . . . , xn)tha mn iu kin (2.11).Khi , vi mi cp b s n iu gim(x1, x2, . . . , xn), (y1, y2, . . . , yn), thamn iu kin Schur (2.10), ta u c

F(x1, x2, . . . , xn) F(y1, y2, . . . , yn). (2.16)Nhn xt 2.3.C th ni rng, nh l 2.8cho ta mt cng c rt mnh thc hin qu trnh lm u v thut ton dn bin chng minh nhiudng bt ng thc phc tp.

Tht vy, t (2.16) ta c ngay H qu.

H qu 2.1(Dn bin tng qut).Gi s choF(x1, x2, . . . , xn)l mt hmlinbin. Khi vi mi b s n iu gimx1, x2, . . . , xnta c

F(x1, x2, . . . , xn) F(y ,y ,x3, . . . , xn) , (2.17)trong ,y= x1+x2

2 .



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Chng minh. Xt hai b sx1, x2, . . . , xnv

x1+x2

2 ,

x1+x2

2 , x3, . . . , xn,ta thy

x1x1+x22

x1+x2x1+x22

+x1+x2

2x1+x2+x3x1+x2

2 +

x1+x22

+x3. . . . . . . . .

x1+x2+ +xn1x1+x2

2 +

x1+x22 +x3+ +xn1

x1+x2+ +xn= x1+x22

+x1+x2

2 +x3+ +xn.

R rng hai b s cho tha mn iu kin Schur (2.10). Khi , p dngnh l2.8ta c iu phi chng minh.

iu kin (2.17) c th bin i thnh mt s dng khc nh

F(x1, x2, . . . , xn) F(x1x2, x1x2, x3, . . . , xn),

F(x1, x2, . . . , xn) F

21x1

+ 1x2

, 21x1

+ 1x2

, x3, . . . , xn

,

F(x1, x2, . . . , xn) F

x21+x22

2 ,

x21+x

22

2 , x3, . . . , xn

.

Ngoi ra, cn nhiu dng khc na ty theo yu cu ca bi ton.Sau y, chng ti trch dn mt vi v d s dng phng php dn bin

tng qut.

V d 2.1 (Bt ng thc AM-GM). Chng minh rng vi mi s thcdngx1, x2, . . . , xnta lun c

x1+x2+ +xn n nx1x2 . . . xn.Chng minh. tf(x1, x2, . . . , xn) =x1+ x2+ + xn n nx1x2 . . . xn.

Theo (1.2) ta c

x1+x2 2x1x2= x1x2+ x1x2



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(

x1

x2)

2

0

suy raf(x1, x2, . . . , xn) f(x1x2, x1x2, x3, . . . , xn).

Do f(x1, x2, . . . , xn) f(t , t , . . . , t) = 0trong t= nx1.x2. . . . xn.Nh vy, t tng chnh ca phng php ny l lm gim dn s bin

trong bt ng thc, dn a bt ng thc v dng n gin hn, c thgii quyt bi ton. Ta c th gi ngn gn cho cch hiu ton b phng

php ny l gim bin.Sau y, chng ti trnh by cc kt qu v dn bin ca mt s tc gi,

c tham kho t[4].

2.3.2 Mt s nh l v dn bin

Khi ni v cc nh l v dn bin, ta ngh ti 2 kt qu u tin ht scn tng, l nh l dn bin mnh (SMV1), v nh l dn bin khng xc

nh (UMV2). Trong mt s trng hp, chng ta s s dng cc nh l dnbin ny.

B 2.1(Theo[4]).Gi sx1, x2, . . . , xnl mt b s thc ty . Ta thchin lin tip php bin i sau.

1. Chnj, j {1, 2, . . . , n}l hai ch s sao cho

xi = min{x1, x2, . . . , xn}, xj = max{x1, x2, . . . , xn}.

2. Thayxivxj bixi+xj

2 (nhng vn gi ng th t ca chng trong

dy s).

Khi sau v hn ln thc hin bin i ni trn th mi s aiu tin tigii hn

x=x1+x2+ +xn

n .

1

Stronger Mixing Variable - Dn bin mnh - ca Phm Kim Hng2Undefined Mixing Variable - dn bin khng xc nh - ca inh Ngc An



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36

Ta gi php bin i B 2.1l . T B 2.1, ta suy ra c kt

qu trc tip sau, c tn l "Dn bin mnh". Dn bin mnh s l 1 cngc c lc gip ta gii quyt c cc bi ton bt ng thc cc tr tc ti tm (cc gi tr bng nhau).

nh l 2.9 (Dn bin mnh SMV (Theo [4])). Nu f : I Rk R,I= [, ] [, ] [, ]vi , Rl hm lin tc i xng v bchn di tha mn iu kin

f(x1, x2, . . . , xn) f(y1, y2, . . . , yn)viy1, y2, . . . , ynl b s thu c t b s x1, x2, . . . , xntheo php bin i, th ta c f(x1, x2, . . . , xn) f(x , x , . . . , x)vi x= x1, x2, . . . , xn

n .

Khi s dng nh l SMV dn bin th ta ch cn chn ra s nh nhtv s ln nht. ng dng hiu qu nht ca nh l SMV l chng minh btng thc 4 bin.

Php bin i c th khc hn, chng hn thay thnh

xy,

21x

+ 1y

,

x2 +y2

2 ,

hoc bt k mt dng trung bnh no khc. Ty theo gi thit ca bi tonm ta cn chn cch dn bin cho ph hp.

V d 2.2 (IMO Shortlist 1997, Vit Nam).Gi sa,b,c,dl cc s thc

khng m c tng bng 1. Hy chng minh bt ng thcabc+bcd+cda+dab 1

27+

176

27abcd.

Chng minh. Khng mt tnh tng qut ca bi ton ta gi sabc d. Xt

f(a,b,c,d) =abc+bcd+cda+dab 17627

abcd,

f(a,b,c,d) =ac(b+d) +bd

a+c 17627

ac

.



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37

T gi thit suy ra

a+c 12

(a+b+c+d) =1

2.

Do

1

a+

1

c 4

a+c 8 176

27 f(a,b,c,d) f

a,

b+d

2 , c,

b+d

2

.

Ta ch xt php bin ivi(b,c,d)v theo kt qu chng minh

f(a,b,c,d) f(a,t,t,t), t= b+c+d3

.

Cui cng ta ch cn chng minh nu a+ 3t= 1th

3at2 +t3 127

+176

27at3.

Ta c th chng minh iu ny, khi thay a = 1 3tta c bt ng thc hin

nhin sau (1 3t)(4t 1)2(11t+ 1) 0.ng thc xy ra khi v ch khia= b = c = d =

1

4hoca= b = c =

1

3, d= 0

hoc cc hon v tng ng.Bi ton c chng minh [4].

K tip chng ti trch dn thm mt kt qu c tn "dn bin khng xcnh". Dn bin khng xc nh i hi gi thit t ln 2 bin bt k, tuy

nhin n cho php ta dung ha c 2 trng hp cc tr t c ti tm (ccgi tr bng nhau) v ti bin (c 1 gi tr bng 0) di mt dng tng qut.

nh l 2.10 (Dn bin khng xc nh UMV (Theo [4])). Cho f l mthm lin tc i xng xc nh trn tp U( Rn) Rtha mn iu kin

f(. . . , x , . . . , y , . . . )

minf(. . . ,x+y2 , . . . ,x+y2 , . . . ), f(. . . , 0, . . . , x+y, . . . ) .S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn


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Khi vi mi b (x1, x2, . . . , xn)

Uth

f(x1, x2, . . . , xn) min{Ct}n1t=0 .Trong Ctl gi tr ca hm fkhi c ts bng 0 v cc s cn li bngnhau. Ni cch khc, gi tr nh nht ca biu thc f(x1, x2, . . . , xn)s tc khi v ch khi trong cc s x1, x2 . . . , xnc ts bng 0, cc s cn libng nhau. y tl mt gi tr nguyn no trong{0, 1, . . . , n 1}.

Phm vi ng dng hiu qu ca nh l UMV l chng minh bt ng

thcnbin.V d 2.3(IMO Shorrtlist).Choa, b, c, dl cc s thc khng m c tngbng 1. Chng minh rng

abc+bcd+cda+dab 127

+176

27abcd.

Chng minh. Xt

f(a,b,c,d) =abc+bcd+cda+dab

176

27

abcd

=ab(c+d) +cd(a+b) 17627

abcd.

D thy

f(a,b,c,d) f

a+b

2 ,

a+b

2 , c , d

(c + d)

ab (a+b)2

4

+

176

27cd

(a+b)2

4 ab

0 176

27cd c + d.

f(a,b,c,d)

f(0, a+b, c, d)

ab(c+d 17627

abcd) 0 c+d 17627

cd.

T cc chng minh trn suy ra

f(a,b,c,d) max

f

a+b

2 ,

a+b

2 , c , d

, f(0, a+b, c, d)

.

Theo nh l UMV, suy ramax f(a,b,c,d)t c khi trong cc sa, b, c, dc mt s bng 0 hoca = b = c = d = 1. T ta c iu phi chng minh

[4].



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Chng 3

Mt s p dng

Trong chng ny chng ti trch dn nhng bi ton bt ng thc pdng phng php dn bin.

Cc vn trnh by trong chng ny c chng ti tham kho v trchdn ch yu t mt s ti liu[3], [8], [4].

3.1 Mt s k thut thng dng trong gii bi ton

bt ng thcTrc khi i vo nhng bi ton p dng, chng ti trch dn mt s k

thut hay dng trong gii bi ton bt ng thc s dng phng php dnbin, nhm mc ch gip cho li gii bi ton ngn gn hn, v c th ngin hn.

3.1.1 K thut chun ha

i vi mt s bi ton bt ng thc dng thun nht ng bc, mtphng php hiu qu gii l ta c th chun ha theo nhiu cch khcnhau ty thuc vo c im ca tng bi ton. Vic chun ha thch hps tng thm gi thit ca bi ton, t ta c cc bc bin i hoc nhgi c li gii gn gng v p .

Cc bi ton bt ng thc thng gp a phn c ba bin. Do dtip cn vi k thut chun ha, chng ta s xem xt k thut ny trong bi

ton bt ng thc c ba bin.



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40

Hm thun nht Hmfgi l thun nht cpk trn min xc nh D nuf(ta,tb,tc) =tkf(a,b,c), (a,b,c) D3.

K thut chun ha bt ng thc Gi s ta cn chng minh bt ng thcf(a,b,c) 0. Khi ta c: f(ta,tb,tc) =tkf(a,b,c) 0vi t >0.

Nh vy, ta c th gi s rnga, b, ctha mn mt iu kin no (ccho php trong minD) v chng minh bt ng thc trong trng hp ny.Gi s ta chng minh c bt ng vi b ba s (x,y,z) D3, khi vi

mi(a,b,c) D3

, t a= tx, b= ty, c= tz, t >0 (lun tn ti tnh th).Khi f(a,b,c) =tkf(x,y,z) 0v c iu phi chng minh.Cn mt dng bt ng thc thun nht khc, l g(a,b,c) 0, trong

g(ta,tb,tc) =g(a,b,c), t D.

3.1.2 K thut sp th t b s

Trong qu trnh gii bi ton bt ng thc, vic gi sa= min{a,b,c}

l mt k thut rt hay c p dng dn bin. Nu bt ng thc ba bini xng th ta c th gi sa b c(hoc a b c), cn trong trnghp bt ng thc ba bin hon v th ta c th gi sa = min{a,b,c} (hoca= max{a,b,c}).

3.2 K thut dn bin

Chng ta sp xp cc cp gi tr c cc cp gi tr u nhau, c lm

nh vy chng ta s dn a tt c cc gi tr u nhau (a tt c cc binbng nhau), hoc a mt s cp gi tr u nhau (a mt s cp gi trbng nhau).

n gin v d tip cn hn, chng ti s trnh by nhng m t cachng ti v dng bi ton bt ng thc 3 bin. Bi ton bt ng thc 3bin c dng f(x,y,z) 0, vi x, y, zl cc bin s thc tha mn nhngtnh cht nht nh.



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41

3.2.1 Dn cc bin bng nhau

Mc tiu ca chng ta l a bi ton bt ng thc v trng hp dubng xy ra khi tt c cc bin u bng nhau (ta gi l cc tr t ti tm),hoc trng hp du bng xy ra khi mt s cc bin bng nhau (ta gi lcc tr t c c tnh i xng).

Chng ti a bi ton v dng f(x,y,z) f(t,t,z ), vitl gi tr phhp ty theo mi lin h gia x, y, z(ta gi l dn hai bin bng nhau). Sau ta c bi ton dng f(t,t,z )

0. Tip tc lm theo phng php trn

th s c bi ton dng f(u,u,u) 0, y l dng bt ng thc mtbin, d dng kim tra hon tt chng minh.

Chng ta s n vi bi ton p dng u tin l mt bt ng thc rtquen thuc trong chng trnh ton trung hc, bt ng thc Nesbitt.

Bi ton 3.1 (Bt ng thc Nesbitt). Chng minh rng vi mi a,b,cdng ta c

a

b+c+

b

a+c+

c

a+b3

2.

Chng minh. tf(a,b,c) = a

b+c+

b

a+c+

c

a+bvt =

a+b

2 . Khi

d thy

f(a,b,c) =a2 +b2 +c(a+b)

ab+c2 +c(a+b)+

c

a+b 2t

2 + 2tc

t2 +c2 + 2ct+

c

2t=f(t,t,c).

Do

f(a,b,c) g(t, t, t) =3

2 vit=a+b+c

3 .

Bi ton c chng minh [4].

Bi ton 3.2(AM-GM vi 3 bin).Cho x, y, zkhng m, chng minh rng

3

xyzx+y+z3

. (3.1)

Chng minh. Do bt ng thc (3.1) ng bc nn bng cch chun hata c th gi s

x+y+z= 1. (3.2)



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42

Vit li bi ton di dngf(x,y,z)

0vif(x,y,z) = 1

27xyz. Ta thy

rng khi thay xvybi t= x+y2

th iu kin (3.2) vn bo ton (tc l

vn ct+t+z= 1), nn ta ch phi xem xt s thay i caxyz.Theo bt ng thc AM-GM tht2 xy, nnt2z xyz. Vyf(x,y,z)

f(t,t,z ).Cui cng l z= 1 2tnn ta c

f(t,t,z ) = 1 27t2z= 1 27t2(1 2t) = (1 + 6t)(1 3t)2 0.

Bi ton c chng minh. ng thc xy ra khi x= y v3t= 1, nghalx= y =

1

3, tng ng vix= y =z.

Bi ton c chng minh.

Bi ton 3.3(Bt ng thc Schur).Cho a, b, c 0, chng minh rnga3 +b3 +c3 + 3abc a2(b+c) +b2(a+c) +c2(a+b).

Chng minh. Xtf(a,b,c) =a3 + b3 + c3 + 3abc

a2(b + c)

b2(a + c)

c2(a+b). t t= b+c2

, ta hy vng cf(a,b,c) f(a,t,t). Xt

d= f(a,b,c) f(a,t,t) =

b+c 54

a

(b c)2.

Ta thy vi a, b, cl cc s khng m ty th khng chc c d 0. Tuynhin, nu gi sa = min{a,b,c}th ta vn c d 0. Khi ta ch cnphi chng minh f(a,t,t) 0. Nhng bt ng thc ny tng ng vi

a(a t)2

0nn hin nhin ng. Vy bi ton c chng minh.Bi ton 3.4 (Gabriel Dospinescu). Chng minh rng nu a, b, c > 0saochoa2 +b2 +c2 = 3, th

a3(b+c) +b3(c+a) +c3(a+b) 6.Chng minh. t f(a,b,c) = a3(b+c) +b3(c+ a) +c3(a+ b), v t

t= b2 +c2

2 . Ta c

f(a,b,c) f(a,t,t) = a3(2t b c) +a(b3 +c3 2t3) +t2(2bc 2t2).



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43

Gi sa= min{

a,b,c

}. Th

b3 +c3 2t3 = (b+c)(b2 bc+c2) 2t3 2t(b2 bc2c) 2t3 =t(b c)2.Nh vy, ta c bt ng thc

a3(2t b c) +at(b c)2 t2(b c)2 0,vi mi2t b c 0va t. Do , ta cn chng minh rng a2 + 2t2 = 3,thf(a,t,t) 6. Ta c th vit

a3t+at3 +t4 3 at(3 t2) 3 t4

(3 2t2)t2(3 t2)2 (3 t4)2

iu ny tng ng vi

3(t2 1)2(t4 3t2 + 3) 0.Bi ton c chng minh [3].

Bi ton 3.5(Kwant).Cho x, y,z >0. Chng minh rng

2(a2 +b2 +c2) + 3 3

a2b2c2 (a+b+c)2.Chng minh. t f(a,b,c) = 2(a2 +b2 +c2) + 3 3

a2b2c2 (a+b+c)2.

Tfta nhn thy c th dn hai bin theo trung bnh nhn ( 3

a2b2c2 khngi). Ta c

f(a,b,c) f(a,

bc,

bc) = 2(b

2

+c

2

2bc) + (a+ 2

bc)

2

(a+b+c)2

= (

b

c)2[(

b+

c)2 2a].Nh vy, nu a= min{a,b,c}(ta c th hy vng gi thit ) th ta c

f(a,b,c) f(a,

bc,

bc).

Do , vic chng minh bt ng thc ban u, thy rngf(a,b,b) 0vi mia, b. Nhng bt ng thc ny tng ng vi

2(a2 + 2b2) + 3 3a2b4 (a+ 2b)2,



44/53

44

haya2 + 3 3a2b4 4ab.

iu ny ng bi bt ng thc AM-GM vi a2, 3

a2b4, 3

a2b4, 3

a2b4.Bi ton c chng minh [3].

Bi ton 3.6 (Gabriel Dospinescu, Mircea Lascu, Marian Tetiva). Chngminh rng vi nhng s thc dng bt k a, b, c, ta c

a2 +b2 +c2 + 2abc+ 3

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

Chng minh. tf(a,b,c) =a2+b2+c2+abc+2abcabbcca.Ta cn chng minh rng tt c cc gi tr cafu khng m. Nua, b, c >3

th ta hin nhin c 1

a+

1

b+

1

c a2 +b2 +c2 + 2 a b c >0.

Do , chng ta c th gi s rng a 3v t m = b+c2

. D dng tnh

cf(a,b,c) f(a,m,m) =(3 a)(b c)

2

4 0,

v do c th chng minh c rng f(a,m,m) 0, tc l(a+ 1)m2 2(a+ 1)m+a2 a+ 2 0. (3.3)

R rng bt ng thc (3.3) ng, bi v bit thc ca phng trnh bchai t (3.3) l = 4(a+ 1)(a 1)2 0. Bi ton c chng minh [8].

By gi, chng ti trch dn mt bi ton bt ng thc 4 bin s dngphng php dn bin mnh SMV chng minh nh sau.

Bi ton 3.7(Bt ng thc Tukervici).Chng minh rng vi mi s thcdnga, b, c, dta c

a4 +b4 +c4 +d4 + 2abcd a2b2 +b2c2 +c2d2 +d2a2 +a2c2 +b2d2.Chng minh. Gi sa

b

c

d. Xt

f(a,b,c,d) =a4 + b4 + c4 +d4 +2abcda2b2b2c2c2d2d2a2a2c2b2d2



45/53

45

=a4 +b4 +c4 +d4 + 2abcd

a2c2

b2d2

(a2 +c2)(b2 +d2).

Suy ra

f(a,b,c,d) f(ac, b,ac,d) = (a2 c2) (b2 +d2)(a c)2 0.

Do theo nh l SMV, xt vi php bin i ca (a,b,c), ta ch cnchng minh bt ng thc khi a = b = c = t d. Bt ng thc lc nytng ng vi

3t4 +d4 + 2t3d 3t4 + 3t2d2 d4 +t3d+t3d 3t2d2.Hin nhin ng theo bt ng thc AM-GM. ng thc xy ra khi v chchia= b = c = d hoca= b = c, d= 0hoc cc hon v.

Bi ton c chng minh [4].Tip theo, chng ti trnh by thm mt trng hp p dng phng

php dn bin sau y.

3.2.2 Dn bin ra binThng thng cc bi ton bt ng thc thng s dng phng php

dn bin u (dn cc bin bng nhau) th chiu ngc li ta phi s dngdn bin ra bin.

Gi s, xt bt ng thc f(x,y,z) 0vix, y, z 0, ta c th hy vngvo nh gif(x,y,z) f(0, s , t), trong s, tl cc i lng ph hp tcc bin x, y, z. Ta c gng chn s, tsao cho hiu d = f(x,y,z) f(0, s , t)n gin v d dng nh gi. Cui cng ta ch vic kim tra f(0, s , t) 0.

Trong Bi ton3.3, ta thy cch chng minh bt ng thc ny bngcch dn 2 bin bng nhau. Tuy nhin, ngoi im a = b = c, ng thc cnxy ra khi a = b, c = 0(v cc hon v). Do , ta c th chng minh btng thc ny nh sau.

Bi ton 3.8(Bt ng thc Schur).Cho a, b, c 0. Chng minh rng

a

3

+b

3

+c

3

+ 3abc a2

(b+c) +b

2

(c+a) +c

2

(a+b).



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46

Chng minh. tf(a,b,c) =a3 + b3 + c3 + 3abc

a2(b + c)

b2(c + a)

c2(a + b). Ta hy vng s cf(a,b,c) f(0, s , t)vis = a+b2

, t= a+c2

. Ta

xt hiu

da=f(a,b,c) f

0,a+b

2 ,

a+c

2

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

Tng t ta cng c cc hiu

db=f(a,b,c)

f0,b+a

2 ,

b+c

2 =b(a+b 2c)(b+c 2a),

dc=f(a,b,c) f

0,c+a

2 ,

c+b

2

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

Do tnh i xng nn ta gi sda= max{da, db, dc}. Khi nuda dadbdc=abc(a+b 2c)2(b+c 2a)2(c+a 2b)2.

iu ny l mu thun vi gi thit a, b, c 0.Vyda 0nn f(a,b,c) f(0, s , t)vi s= a+b2 , t= a+c2 . Cui cng

ta thy

f(0, s , t) =t3 +s3 t2s ts2 = (t+s)(t s)2 0.Chng minh c hon tt.

Sau y, l mt bi ton m khng th p dng dn bin a tt ccc bin bng nhau, m ch c th p dng dn bin ra bin.

Bi ton 3.9(Hojoo Lee).Cho a, b, c 0,ab+bc+ca= 1. (3.4)

Chng minh rng1

a+b+

1

b+c+

1

c+a 5

2. (3.5)

Chng minh. D thy ng thc khng th xy ra khia=b =c. Ta cth nhm c ng thc xy ra khi a = b = 1, c = 0v cc hon v. Khi ta chng minh bt ng thc theo hng a mt gi tr bng 0.



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47

Xtc= 0, khi ta cn chng minh1

a+

1

b+

1

a+b 5

2, via, b= 1. (3.6)

ts= a+bth bt ng thc (3.6) tng ng vi

s+1

s 5

2,

hay(2s 1)(s 2) 0.

Bt ng thc ny l hin nhin v s = a+b 2

ab = 2. Vy cng vicca ta by gi l dn mt bin v 0.

tf(a,b,c) = 1

a+b+

1

b+c+

1

c+a. Ta hy vng

f(a,b,c) f

a+b, 1

a+b, 0

(m bo iu kin (3.4)). Ta xt hiu

d= f(a,b,c)

fa+b, 1

a+b

, 0=

1

a+b+

1

a+ 1aba+b

+ 1

b+ 1aba+b

1

a+b+a+b+

1

a+b+ 1a+b

= 1

1 +a2+

1

1 +b2 1 1

1 + (a+b)2.

Quy ng ln ta thy d 0nu 2(1 ab) ab(a+b)2.Gi sc= max{a,b,c}th 2(1 ab) = 2c(a+b)ab(a+b)2. Vy lc

nyd 0. Bi ton c chng minh.Nhn xt 3.1.Quan st phng php dn bin trnh by Mc3.2.1,Mc3.2.2, ta thy rng, chng t f(x,y,z) f(t,t,z ), ta ch vic xthiud= f(x,y,z) f(t,t,z ), ri tm cch nh gi sao cho d 0. Phngphp ny t ra rt hiu qu trong vic chng minh cc bi ton bt ngthc c in, v n thin v bin i i s. Tuy nhin, c rt nhiu bi tonbt ng thc nu s dng phng php ny li cho ra nhng biu thc khphc tp v vic chng minh mt bi ton bt ng thc tr nn kh cngknh, chng hn nh hmf(x,y,z) =xk +yk +zk.



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48

3.2.3 Dn bin trong lp hm li, lm

Hm li c th ng dng trong rt nhiu bt ng thc c in. Ta trnh by v dn bin trong lp hm li, lm ti chng 2. Tuy nhin, n gin v d tip cn hn, chng ta s trnh by li cch p dng dn bintrong lp hm li, lm vi bt ng thc 3 bin.

Cho hai b s x1, x2, x3vy1, y2, y3tha mn

x1 x2 x3vy1 y2 y3v x1 y1

x1+x2 y1+y2x1+x2+x3=y1+y2+y3.

Khi , vi hm s f(x)tha mn

f(x) 0, x (a, b)v

f(x1) +f(x2) +f(x3) f(y1) +f(y2) +f(y3).Ta lun c

f(x1) +f(x2) +f(x3) 2f

x1+x22

+f(x3). (3.7)

Nhn vo (3.7), ta c th thy mc tiu ca chng ta l a

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

v mt hm mt bin gi s nh

3f

x1+x2+x3

n

.

Sau y, chng ti trnh by mt s bi ton p dng ca phng phpdn bin trong lp hm li, lm.

V d 3.1(Bt ng thc AM-GM).Chox1, x2, . . . , xnl cc s thc khngm,n 1. Chng minh rng

nx1x2 . . . xnx1+x2+ +xnn

. (3.8)



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49

Chng minh. Ly logarit hai v, ta chuyn bt ng thc (3.8) v dng

ln

x1+x2+ +xn

n

ln(x1) + ln(x2) + + ln(xn)

n. .

Hm s f(x) = ln(x) i tR+ R kh vi hai ln v f(x) =x2 0. Do , hm g(x) = f(x)s tho mng(x)> 0, x >0. Vyg l hm li. T , p dng bt ng thc Jensen ta c iu phi chngminh.

Trn thc t, vic s dng hm li dn bin khng c nhiu bi tonp dng cho ngay kt qu, tuy nhin n c th gip gii quyt 1 trnghp no ca bi ton, cc trng hp cn li ta c th chng minh bngcch khc. V d sau s cho thy iu ny.

Bi ton 3.10.Cho cc s thc x, y, zc tng bng 1. Chng minh rng

x

1 +x2+

y

1 +y2+

z

1 +z2 9

10. (3.9)

Chng minh. Xtf(t) = t

1 +t2th bt ng thc (3.9) trng ng

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

x+y+z

3

.

Do , nufl hm li th coi nh bi ton c gii quyt.Ta c

f(t) =2t(3

t2)

(1 +t2)3 ,

nnf(t) 0, t [0, 3]. Vy nu x, y, z [0, 3]th bi ton cgii quyt.

Trong trng hp cn li th chc chn ta s c du bt ng thc thcs. Do vy c vic chia thnh nhiu trng hp con xt.

C th gi sx y zlu x + y+ z= 1vx, y, z / [0, 1]nn zphim. Suy ra f(z)< 0.



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Nuy 0 vf(y)< 0, ta c

f(x) +f(y) +f(z)< f(x)0 v lu f(y), f(x)nghch bin trn[3, +]do

f(x) +f(y) +f(z)< f(x) +f(y)< f(

3) +f(

3)< 9

10.

Bi ton c chng minh.

3.2.4 Dn bin trong lp hm n iu

Gi s, chng minh f(x,y,z) f(t,t,z )vit = x+y2

, ta xt hm s

g(s) =f(t+s, t s, z)vi s 0. Sau chng minh gtng vi s 0(tathng dng o hm). Suy ra g(s) g(0), s 0, sau ta s c iucn chng minh.

Sau y, chng ti trnh by mt bi ton p dng minh ha cho dn bintrong lp hm n iu.

Bi ton 3.11 (VMO 2002). Chng minh rng vi nhng s thc ty x,y,zsao cho

x2 +y2 +z2 = 9,

th bt ng thc2(x+y+z)

xyz+ 10

ng.Chng minh. Chof(x,y,z) = 2(x+y+z)xyz. Ta thy rngf(x,y,z) 10khi x2 +y2 +z2 = 9.

B iu kinx2 + y2 + z2 = 9nh c, ta s dn 2 bin v dng trung bnhbc hai.

Xem xt

f

x,y2 +z2

2 ,y2 +z2

2 f(x,y,z) =



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51

= 2x+ 2y2 +z2

2 xy2 +z2

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

= 2(y z) 2xy z2

= 2(y z)

2

y+z x

2

.

Nux, y, z >0, ta xt 2 trng hpTrng hp th nht:0 x y z, th

2(x+y+z)

xyz

23(x2 +y2 +z2)

1 = 6

3

1

10.

Trng hp th 2: 0< x 1, th2(x+y+z) xyz


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52

Kt lun

Phng php gim bin v dn bin l mt trong cc hng ang pht

trin mnh m trong phng php chng minh bt ng thc ni ring, vca ngnh phng php ton s cp ni chung. S pht trin vt bc nychnh l nh s pht hin v nh hng lm gim s bin n gin havic chng minh bi ton bt ng thc.

Lun vn trnh by c mt s vn nh sau:

1. Trnh by cc kin thc c s v mt s bt ng thc c bn lin quann gi tr trung bnh.

2. Trnh by cc kin thc v tam thc bc hai, t trnh by phngphp gim bin trong bt ng thc i s.

3. Trnh by kin thc c s v gn u, hm li, t trnh by phngphp dn bin tng qut.

4. Ttrch dn mt s kt qu v dn bin ca mt s tc gi.

5. Trnh by mt s p dng ca phng php dn bin.



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Ti liu tham kho

[1] Nguyn Vn Mu, 2006, Bt ng thc, nh l v p dng,NXB GD.

[2] Nguyn Vn Mu, 2007, Cc bi ton ni suy v p dng, NXB GD.

[3] Nguyn Vn Mu (Ch bin), 2009,Bt ng thc v mt s vn linquan,NXBGD.

[4] Phm Kim Hng, 2007,Sng to bt ng thc, NXB Tri thc.

[5] J.Michael Steele, 2004,Cauchy Schwarz master class,Cambridge.

[6] V.Cirtoaje, 2002, On Popovicius Inequality for Convex Function,Gazeta Matematica A, 4(2002), 247-253.

[7] M.S.Klamkin, 2000-2003, On a "Problem of the Month", Crux Mathe-maticorum with Mathematical Mayhem.

[8] Mitrinovic D.S., Pecaric J.E., Fink A.M., 1993, Classical and New In-equalities in Analysis,Cluwer Academic Publishers.

[9] A.Lupas, 1982, On an Inequality for Convex Function,Gazeta Matem-atica A, (1982), 1-2.

[10] T. Popoviciu, 1965, Sur certaines inegalites qui caracterisent les func-tions convex,Analele Stiintifice ale Univ, Iasi, Sectia Mat. 11B (1965),155-164.

[11] V.Cirtoaje, 1996, Amer. Math. Monthly 103 (1996) 427-428.


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