Skkn Don Bien

8/3/2019 Skkn Don Bien

1/43

Gio vin: Tr n nh Hi n - Tr ng THPT ng Thc H a - Ngh An

M C L C

M c l c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1L i ni u. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Chng 1.Gi tr nh nh t, gi tr l n nh t c a hm s. . . . . . . . . . . . . 4

1.1.M t s ki n th c c s v o hm. . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.Gi tr nh nh t, gi tr l n nh t c a hm s. . . . . . . . . . . . . . . . . 5

1.3.M t s v d tm gi tr nh nh t, gi tr l n nh t c a hm6Chng 2.K thu t gi m bi n trong bi ton tm gi tr nh nh t, g

l n nh t c a bi u th c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.Bi ton tm gi tr nh nh t, gi tr l n nh t c a bi u th c

phng php th. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2.Bi ton tm gi tr nh nh t, gi tr l n nh t c a bi u thx ng. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3.Bi ton tm gi tr nh nh t, gi tr l n nh t c a bi u th

hi n tnh ng c p. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.4.Bi ton tm gi tr nh nh t, gi tr l n nh t c a bi u th c

ba bi n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30K t lu n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Ti li u tham kh o. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

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L I NI U

Trong sch gio khoa l p 12 Gi i tch trnh by cch tm gi tr l ntr nh nh t c a hm s . V v y m t s d ng bi ton tm gi tr nh nl n nh t c a m t bi u th c ch a m t bi n tr nn n gi n. Bi ton nh nh t, gi tr l n nh t l m t bi ton b t ng th c v y l m t trod ng ton kh chng trnh trung h c ph thng. Trong cc bi ton tl n nh t, gi tr nh nh t c a m t bi u th c dnh cho h c sinh kh, gth c c n tm gi tr nh nh t, gi tr l n nh t th ng ch a khng t h

Khng nh ng th , cc bi ton kh th ng c gi thi t rng bu c gi a Vi c chuy n bi ton tm gi tr nh nh t, gi tr l n nh t c a m t

khng t hn hai bi n sang bi ton tm gi tr nh nh t, gi tr l n nhs ch a m t bi n s gip chng ta gi i c bi ton tm gi tr nh nl n nh t c a m t bi u th c. V n t ra l nh ng d ng bi ton tm nh t, gi tr l n nh t no th chuy n v c d ng bi ton tm gi trgi tr nh nh t c a hm s ch a m t n. V v y chng ti ch n ti

" ng d ng o hm tm gi nh nh t, gi tr nh nh t c a mth c".

Trong qu trnh gi ng d y, b i d ng h c sinh gi i v n thi i h cb n thn c rt c m t s kinh nghi m. V v y trong bi vi t nytrnh by chi ti t m t s d ng bi ton tm gi tr nh nh t, gi tr l nm t bi u th c ch a hai bi n m i u ki n rng bu c c a hai bi n ho

th hi n tnh i x ng ho c tnh ng c p, trnh by m t s bi ton tnh nh t, gi tr l n nh t c a m t bi u th c ch a ba bi n m b ng ccchng ta th c hai bi n qua bi n cn l i.

V i m c ch nh v y, ngoi l i m u, m c l c v ph n ti li u thavi t c trnh by trong hai chng.

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Chng 1. Gi tr nh nh t, gi tr l n nh t c a hm s. Trong chngny, chng ti trnh by cc ki n th c c s c n thi t gi i bi ton tnh nh t, gi tr l n nh t c a hm s . cu i chng, chng ti a rad minh ho .

Chng 2. K thu t gi m bi n trong bi ton tm gi tr nh nhtr l n nh t c a bi u th c. Trong chng ny, chng ti trnh by chi ti td ng ton tm gi tr nh nh t, gi tr l n nh t c a m t bi u th c chm i u ki n rng bu c c a hai bi n ho c bi u th c th hi n tnh itnh ng c p, trnh by m t s d ng ton tm gi tr nh nh t, gi trc a m t bi u th c ch a ba bi n b ng cch t n ph ho c th hai bibi n cn l i.

Tc gi xin g i l i c m n chn thnh t i cc th y gio trong t Tonem h c sinh l p 12A-K30, 10C1-K33 tr ng THPT ng Thc H a gip tc gi trong su t qu trnh nghin c u v hon thi n bi bi t.

Trong qu trnh th c hi n bi vi t ny, m c d r t c g ng nhng ktrnh kh i nh ng h n ch , thi u st. Tc gi r t mong nh n c nh

ng gp c a qu th y c, cc b n v cc em h c sinh bi vi t chn.

Xin trn tr ng c m n!

Thanh Chng, thng 05 nm 2011Tc gi

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CHNG 1

GI TR NH NH T, GI TR L N NH T C A HM S

1.1. M t s ki n th c c s v o hm

Trong m c ny chng ti trnh by l i m t s ki n th c v o hm cng th c v o hm.

1.1.1 nh l.N u hai hm s u = u(x) v v = v(x) c o hm trn J th 1. (u + v) = u + v ;

2. (u v) = u v ;3. (uv ) = u v + uv ;

4. (ku ) = ku ;

5. ( uv ) = u vuvv2 , v i v(x) = 0

1.1.2 nh l. o hm c a m t s hm s th ng g p.

(c) = 0 (c l h ng s )(x) = 1

(xn ) = nx n1(nR ) (un ) = nu n1u

( 1x ) = 1x 2 ( 1u ) = uu 2(x ) = 12x (x > 0) (u) = u2u

(ex ) = ex (eu ) = eu u

(ln x) = 1x

(x > 0) (ln u) = uu

(sin x) = cos x (sin u ) = u cos u

(cos x) = sin x (cos u) = u sin u(tan x) = 1 + tan 2 x(x = 2 + k ) (tan u) = u (1 + tan

2 u )

(cot x) = (1 + cot 2 x)(x = k ) (cot u) = u (1 + cot 2 u)4


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1.1.3 Nh n xt. o hm c a m t s hm phn th c h u t th ng g p1. Cho hm sy = ax + bcx + d v ia.c = 0 , ad cb = 0 . Ta cy = ad cb(cx + d)2 .2. Cho hm sy = ax 2 + bx+ cmx + n v ia.m = 0 . Ta cy = amx

2 +2 anx + bnmc(mx + n )2 .

3. Cho hm sy = ax2 + bx+ c

mx 2 + nx + p v ia.m = 0 . Ta cy =(an mb )x

2 +2( ap mc )x+( bpnc )(mx 2 + nx + p)2 .

1.2. Gi tr nh nh t, gi tr l n nh t c a hm s

Trong m c ny chng ti trnh by l i m t s ki n th c v bi ton tnh nh t, gi tr l n nh t c a hm s .

1.2.1 nh ngha.Gi s hm sf

xc nh trn t p h pDR

.a) N u t n t i m t i mx0 Dsao chof (x) f (x0) v i m ix Dth sM = f (x0) c g i lgi tr l n nh t c a hm sf trnD, k hi u lM = maxx

Df (x).

b) N u t n t i m t i mx0 Dsao chof (x) f (x0) v i m ix Dth sm = f (x0) c g i lgi tr nh nh t c a hm sf trnD, k hi u lm = minx

Df (x).

1.2.2 Nh n xt.Nh v y, mu n ch ng t r ng sM (ho cm ) l gi tr l n nh

(ho c gi tr nh nh t) c a hm sf

trn t p h pDc n ch r:a) f (x) M (ho cf (x) m ) v i m ixD.b) T n t i t nh t m t i mx0Dsao chof (x0) = M (ho cf (x0) = m ).

1.2.3 Nh n xt.Ng i ta ch ng minh c r ng hm s lin t c tro n th t c gi tr nh nh t v gi tr l n nh t trn o n .

Trong nhi u tr ng h p, c th tm gi tr l n nh t v gi tr nh nh

s trn m t o n m khng c n l p b ng bi n thin c a n.Quy t c tm gi tr nh nh t, gi tr l n nh t c a hmf trn o n[a ; b] nhsau:

1. Tm cc i mx 1, x 2,...,x n thu c kho ng(a ; b) m t i f c o hm b ng0ho c khng c o hm.

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2. Tnhf (x1), f (x2),...,f (xn ), f (a ) v f (b).3. So snh cc gi tr tm c.

S l n nh t trong cc gi tr l gi tr l n nh t c af trn o n[a ; b], s nhnh t trong cc gi tr l gi tr nh nh t c af trn o n[a ; b].

1.3. M t s v d tm gi tr nh nh t, gi tr l n nh t cs

Trong m c ny chng ti trnh by m t s v d tm gi tr nh nh t, nh t c a hm s .

1.3.1 V d .Tm gi tr nh nh t v gi tr l n nh t c a hm sf (x) = x + 4 x2

Bi lm.T p xc nhD= [2;2], f (x) = 1 x4x 2 , f (x) = 0 x =2. B ng

bi n thint 2 2 2f (t ) + 0

f (t )

2

22 d

d

d 2

T b ng bi n thin ta cminx[2;2]

f (x) = f (2) = 2 v maxx[2;2]

f (x) = f (2) = 22.1.3.2 Bi t p.Tm gi tr nh nh t v gi tr l n nh t c a hm s

f (x) = 1 x2 + 23

(1 x2)2.B n c t gi i.

1.3.3 Bi t p.Tm gi tr nh nh t v gi tr l n nh t c a hm s

f (x) = 5 cos x cos5x v i4 x

4

.

B n c t gi i.

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f (x) =21 x 4 + 1 + x2 + 1 x2 + 3

1 + x2 + 1 x2 + 1.

H ng d n. tt = 1 + x2 + 1 x2, 2 t 2.1.3.5 Bi t p.Tm gi tr nh nh t v gi tr l n nh t c a hm s

f (x) = ( x + 1) 1 x2.B n c t gi i.

1.3.6 Bi t p.Tm gi tr nh nh t v gi tr l n nh t c a hm sf (x) = x6 + 4(1 x2)3, v ix[1;1].

B n c t gi i.


f (x) = sin 2 x x, v ix[2

;2

].

B n c t gi i.


f (x) =2x + 3x2 + 1 .

B n c t gi i.

1.3.9 Bi t p.Tm gi tr nh nh t v gi tr l n nh t c a hm sf (x ) = x( 1 x2 + x).

B n c t gi i.

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CHNG 2

K THU T GI M BI N TRONG BI TON TM GI TRNH NH T, GI TR L N NH T C A BI U TH C

T k t qu c a Chng 1 chng ta th y r ng vi c tm gi tr nh nl n nh t c a hm s kh n gi n.Vi c chuy n bi ton tm gi tr nhtr l n nh t c a m t bi u th c khng t hn hai bi n sang bi ton tm nh t, gi tr l n nh t c a hm s ch a m t bi n s gip chng ta gi i tm gi tr nh nh t, gi tr l n nh t c a m t bi u th c.

2.1. Bi ton tm gi tr nh nh t, gi tr l n nh t c a bib ng phng php th

Trong ph n ny chng ti trnh by m t s d ng bi ton tm gi tr ngi tr l n nh t c a bi u th c ch a hai bi n b ng cch th m t bi n ql i. T xt hm s v tm gi tr nh nh t, gi tr l n nh t c a hm

2.1.1 V d .Chox, y > 0 tho mnx + y = 54 . Tm gi tr nh nh t c a bi u th

P =4x

+1

4y.

Bi lm.T gi thi tx + y = 54 ta cy = 54 x. Khi P = 4x + 154x . Xt hms f (x) = 4x + 154x , x (0;

54). Ta cf (x) = 4x 2 + 4(54x ) 2 . f (x) = 0x = 1 v x =

53

(lo i). Ta c b ng bi n thinx 0 1 5

4f (x) 0 +

f (x)

+ d

d d

5

+

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T b ng bi n thin ta cminx

(0; 54 )f (x) = f (1) = 5 . Do min P = 5 t c khi

x = 1 , y = 14 .

Nh n xt.Bi ton ny c gi i b ng cch th m t bi n qua m t biv xt hm s ch a m t bi n.

2.1.2 V d .Chox, y

R tho mny 0, x 2 + x = y + 12 . Tm gi tr nh nh tgi tr l n nh t c a bi u th cP = xy + x + 2 y + 17 .

Bi lm.T gi thi ty 0, x 2+ x = y+12 ta cy = x2+ x12 vx2+ x12 0 hay4 x 3. Khi P = x3+3 x29x7. Xt hm sf (x) = x3+3 x29x7, x[4;3].

Ta cf (x) = 3( x2

+ 2 x 3), f (x) = 0x = 3 v x = 1 . Ta c b ng bi n thinx 4 3 1 3f (x) + 0 0 +

f (x )

13

20 d

d d

12

20

T b ng bi n thin ta cminx[4;3]f (x) = f (1) = 12, maxx[4;3]f (x) = f (3) = f (3) = 20 .Do min P = 12 t c khix = 1 , y = 10 v max P = 20 t c khix = 3, y = 6 ho cx = 3 , y = 0 .

Nh n xt.Bi ton ny c gi i b ng cch th m t bi n qua m t binhng ph i nh gi bi n cn l i. T tm gi tr nh nh t, gi tr lhm s ch a m t bi n b ch n.

2.1.3 V d .Chox, y > 0 tho mnx + y = 1 . Tm gi tr nh nh t c a bi u tP =

x1 x

+y

1 y.

Bi lm. T gi thi tx,y > 0, x + y = 1 ta cy = 1 x, 0 < x < 1. Khi ta cP = x1x

+ 1xx . Xt hm s f (x) = x1x+ 1xx , f (x) = 2x2(1x )1x

x+12xx ,

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f (x) = 0x =12 . B ng bi n thin

x 0 12 1f (x) 0 +

f (x)

+

d d d 2

+

T b ng bi n thin suy ramin P = minx

(0;1)f (x) = f ( 12 ) = 2 t c khix = y = 12 .

Nh n xt.Bi ton ny c gi i b ng cch th m t bi n qua m t biv s d ng cc gi thi t nh gi bi n cn l i. T tm gi tr nh

tr l n nh t c a hm s ch a m t bi n b ch n.2.1.4 V d .Chox, y > 0 tho mnx + y 4. Tm gi tr nh nh t c a bi u th

P =3x2 + 4

4x+

2 + y3

y2.

Bi lm.Ta cP = 3x 2 +44x + 2y 2 + y. p d ng b t ng th c AM -GM ta cP =3x 2 +4

4x +(2

y 2 +y4 +

y4 )+

y2 3x

2 +44x +3

3

2.y.yy 2 .4.4 +

y2 =

12 (x + y)+(

x4 +

1x )+

32 2+2

x. 14.x +

32 =

92 .

Do min P =92 t c khix = y = 2 .

Nh n xt.Bi ton ny c gi i b ng cch nh gi bi u th cP v c g ngchuy n v m t bi n.

2.1.5 Bi t p.Chox, y[3;2] tho mnx3 + y3 = 2 . Tm gi tr nh nh t, g

tr l n nh t c a bi u th cP = x2 + y2.

B n c t gi i.

2.1.6 Bi t p.Chox, y 0 tho mnx + y = 1 . Tm gi tr nh nh t, gi tr nh t c a bi u th cP = xy+1 + yx +1 .

B n c t gi i.

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2.1.7 Bi t p.Chox,y > 0 tho mnx + y = 1 . Tm gi tr nh nh t c a bith cP = x2 + y2 + 1x 2 + 1y 2 .

B n c t gi i.2.1.8 Bi t p.Chox + y = 1 . Tm gi tr nh nh t c a bi u th c

P = x3 + y3 + 3( x 2 y2) + 3( x + y).

B n c t gi i.

2.1.9 Bi t p.Choa,b,x,y

R tho mn0 < a, b 4, a + b 7 v 2 x 3 y.Tm gi tr nh nh t c a bi u th cP = 2x

2

+ y2

+2 x + yxy (a 2 + b2 ) .H ng d n.Tm gi tr l n nh t c aQ = a2 + b2, Xt hm sg(y) = f (x, y ) v i

ny v x l tham s , tm gi tr nh nh t c ag(y) l h (x). Sau tm gi tr nhnh t c a hm sh (x) v ix[2;3].

2.1.10 Bi t p.Chox, y

R tho mnx3 y. Tm gi tr nh nh t c a bi u t

P =

x2+

y2

8x

+ 16.

H ng d n.N ux > 0 th x6 y2 t xt hm sf (x) = x6 + x2 8x + 16 .N ux 0 th x2 + y2 8x + 16 16 v i m ix 0, x3 y.2.1.11 Bi t p.Chox, y

(0;1) tho mnx + y = 1 . Tm gi tr nh nh t cbi u th cP = xx + yy .

H ng d n.Xt hm s f (x) = xx , x

(0;1). Ch ng minhf (x )+ f (y)2

f ( x + y2 ).

Ta cP = xx + (1 x)1x = f (x) + f (1 x) 2f ( 12) = 2.2.1.12 Bi t p.Chox, y > 0 tho mnx + y = 2 . Ch ng minh r ngxy xx yy .

B n c t gi i.

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2.2. Bi ton tm gi tr nh nh t, gi tr l n nh t c a bi i x ng

Trong ph n ny chng ti trnh by m t s d ng bi ton tm gi tr ngi tr l n nh t c a bi u th c ch a hai bi n m gi thi t ho c bi u th ctnh i x ng. T xt hm s v tm gi tr nh nh t, gi tr l n nhs .

2.2.1 V d .Chox2 + y2 = x + y. Tm gi tr nh nh t, gi tr l n nh t c ath cP = x3 + y3 + x2y + xy 2.

Bi lm. tt = x + y, t gi thi tx2+ y2 = x+ y ta c2xy = ( x + y)2(x+ y) = t 2thayxy = t 2 t2 . p d ng b t ng th c(x + y)2 2(x2 + y2) = 2( x + y) hayt2 2t suyra 0 t 2. Khi bi u th cP = ( x + y)3 2xy (x + y) = t 2. Do ta cmax P = 4 t c khit = 2 hayx + y = 2 v xy = 1 suy rax = 1 v y = 1 , ta cmin P = 0 t c khit = 0 hayx = 0 v y = 0 .

Nh n xt.Bi ton ny gi thi t v bi u th cP c cho d i d ng i x

v i hai bi n. V v y, chng ta ngh n cch i bi nt = x + y. Nhng gi i bton tr n v n th ph i tm i u ki n c a bi nt. Sau y l m t s bi ton v i h ng tng t .

2.2.2 V d .Chox, y > 0 tho mnx2 + xy + y2 = 1 . Tm gi tr l n nh t c a bith cP = xyx+ y+1 .

Bi lm. tt = x + y. T gi thi tx, y > 0 vx2 + xy + y2 = 1 suy raxy = t2 1.p d ng b t ng th c(x + y)2 4xy suy ra0 < t 13 . Khi P = t 1 333 .V v ymax P = 333 t c khi(x ; y) = ( 23 ; 13 ) ho c(x ; y) = ( 13 ; 23).2.2.3 V d .Chox, y

R tho mnx + y = 1 v x2 + y2 + xy = x + y + 1 . Tm

gi tr nh nh t, gi tr l n nh t c a bi u th cP = xyx+ y+1 .

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Bi lm. tt = x + y. T gi thi tx2 + y2 + xy = x + y + 1 ta c(x + y)2 xy =(x + y) + 1 hay xy = t2 t 1. p d ng b t ng th c(x + y)2 4xy suy ra3t2

4t

4

0 hay

23

t

2. Khi P = t 2 t1t+1 . Xt hm s f (t) = t

2

t1t+1 ,f (t) = t

2 +2 t(t +1) 2 , f (t) = 0t = 0 v t = 2(lo i). B ng bi n thin

t 23 0 2f (t ) 0 +

f (t )

13

d d

d

1

13

T b ng bi n thin ta cmin P = mint

[23 ;2]

f (t ) = f (0) = 1 t c khi(x ; y) =(1;1) ho c(x ; y) = (1; 1) vmax P = maxt

[

23 ;2]

f (t) = f (23 ) = f (2) = 13 t c khix = y = 13 ho cx = y = 1 .2.2.4 V d .Chox, y

R tho mn0 < x, y 1 v x + y = 4 xy . Tm gi tr nh

nh t, gi tr l n nh t c a bi u th cP = x2 + y2 xy .Bi lm. tt = x + y. T gi thi t0 < x, y

1 v x + y = 4 xy suy raxy = t4 v

1 t 2. Khi P = ( x + y)23xy = t234 t . Xt hm sf (t) = t234 t, f (t) = 2 t 34 ,f (t) = 0t =

38(lo i). B ng bi n thin

t 1 2f (t) +

f (t)

14

52

T b ng bi n thin ta cmin P = mint[1;2]

f (t) = f (1) = 14 t c khix = y = 12v max P = max

t[1;2]

f (t) = f (2) = 52 t c khi(x ; y) = ( 222 ;

2+ 22 ) ho c(x ; y) =

( 2+ 22 ; 222 ).

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2.2.5 V d .Chox, y

R tho mnx, y = 0 vxy (x + y) = x2 + y2 x y + 2 . Tmgi tr l n nh t c a bi u th cP = 1x + 1y .

Bi lm. tt = x + y. T gi thi txy (x + y) = x2

+ y2

x y + 2 hayxy (x + y) =(x + y)2 2xy (x + y) + 2 suy raxy = t2

t +2t +2 . p d ng b t ng th c(x + y)2 4xysuy ra t 3 2t

2 +4 t8t +2 0 hay t < 2 v t 2. Khi P = x+ yxy = t2 +2 t

t 2 t +2. Xt hm s

f (t) = t2 +2 t

t 2 t+2, f (t ) = 3t

2 +4 t+4(t 2 t+2) 2

, f (t ) = 0t = 2 v t = 23(lo i). B ng bi n thint 2 23 2 +f (t) 0 +0

f (t)

1 d

d d

27

2 d

d d 1

T b ng bi n thin ta cmax P = maxt< 2vt 2

f (t ) = f (2) = 2 t c khix = y = 1 .

2.2.6 V d .Chox, y > 0 tho mnxy + x + y = 3 . Ch ng minh r ng3x

y + 1+

3yx + 1

+xy

x + y x2 + y2 +

32

.

Bi lm. t t = x + y t gi thi tx ,y > 0, xy + x + y = 3 , b t ng th c(x + y)2 4xy ta cxy = 3 t, t > 0 vt2+4 t12 0 hayt 2 ho ct 6(lo i). Khi b t ng th c c n ch ng minh tr thnh3(x + y)

2

6xy +3( x+ y)xy +( x+ y)+1 +xy

x+ y (x+ y)22xy + 32hay 3t 2 +9 t184 + 3tt t2 + 2 t 92 t3 t2 + 4 t 12 0(t 2)( t2 + t + 6) 0 lunng

t 2, d u b ng x y ra khit = 2 hayx = y = 1 .2.2.7 V d .Chox,y > 0 tho mnx2 + y2 = 1 . Tm gi tr nh nh t c a bith cP = (1 + x)(1 + 1y ) + (1 + y)(1 + 1x ).

Bi lm. t t = x + y. T gi thi tx,y > 0 v x2 + y2 = 1 suy raxy = t 2 12v t > 1. p d ng b t ng th c(x + y)2 2(x2 + y2) suy ra1 < t 2. Khi P = [1 + ( x + y) + xy ][x+ yxy ] = t

2 + tt1

. Xt hm s f (t) = t2 + tt1

, f (t) = t2

2t1t1,

f (t) = 0t = 1 2(lo i). B ng bi n thin14


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t 1 2f (t) || f (t)

+ d

d d 4 + 32


(1;2]f (t) = f (2) = 4 + 3 2 t c khi

x = y = 12 .

2.2.8 V d .Chox, y > 0 tho mnx + y + 1 = 3 xy . Tm gi tr l n nh t c a bith cP = 3xy(x +1) + 3

yx (y+1) 1x 2 1y 2 .

Bi lm. tt = x + y. T gi thi tx + y + 1 = 3 xy suy raxy = t +13 . p d ngb t ng th c(x + y)2 4xy ta c3t2 4t 4 0 hay t 2 v t 23(lo i). Ta cP = 3xy (x + y)+3( x + y)

2

6xyxy [xy +( x+ y)+1] (x + y) 2 2xyx 2 y 2 =

9(4t 2 t2)4( t+1) 2 3(3 t 2 2t2)(t +1) 2 =

3(5 t+2)4(t +1) 2 . Xt hm s

f (t) = 3(5t+2)4( t+1) 2 , f (t) =3(5t+3)4(t +1) 3 , f (t) = 0t =

35(lo i). B ng bi n thin

t 2 +f (t) f (t)

1 d

d d

0

T b ng bi n thin ta cmax P = maxt[2;+ )

f (t) = f (2) = 1 t c khix = y = 1

2.2.9 V d .Chox, y khng ng th i b ng0 tho mnx + y = 1 . Tm gi tr nhnh t c a bi u th cP = 1x 2 + y2 + x

2

y 2 +1 +y2

x 2 +1 .

Bi lm. tt = x2 + y2, ta c(x + y)2 = 1 suy ra(x2 + y2)+2 xy = 1 hayxy = 1t2 .p d ng b t ng th c(x + y)2 2(x2 + y2) suy rat 12 . Khi

P =1

x2 + y2+

(x4 + y4) + ( x2 + y2)x2y2 + ( x2 + y2) + 1

=1t

+2t2 + 8 t 2t2 + 2 t + 5

.

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Xt hm s f (t ) = 1t + 2t2 +8 t2t 2 +2 t+5 , f (t) = 1t 2 + 4t

2 +24 t +44(t 2 +2 t +5) 2 ,

f (t) = 0(t 1)(t + 1) 2(t 5) = 0 .

Ta c b ng bi n thint 12 1 5 +f (t) 0 + 0

f (t)

125

d d

d 2

125

d d

d 2

T b ng bi n thin ta cmax P = maxt[

12 ;+ ]

f (t) = f ( 12) = f (5) = 12

5 t c khi

(x ; y) = ( 12 ;12 ) ho c(x ; y) = (2; 1);(x ; y) = ( 1;2) v min P = mint

[ 12 ;+ ]

f (t) = f (1) = 2

t c khi(x ; y) = (1; 0); ( x ; y) = (0; 1) .

2.2.10 V d .Chox2 + y2 = 1 . Tm gi tr nh nh t, gi tr l n nh t c ath cP = x1 + y + y1 + x.

Bi lm. tt = x + y, t gi thi tx2 + y2 = 1 hay xy = (x + y)2

12 = t2

12 v b t

ng th cxy (x+ y)2

4 ta c2 t 2. Khi ta cP 2 = ( x + y)2 2xy + 2xy (x + y) + 2xy 1 + ( x + y) + xy = 1 +

t3 t2

+(t2 1)|t + 1 |2 .

N u1 t 2 th P 2 = 12[(1 + 2)t3 + 2t2 (1 + 2)t + 2 2]. Xt hm sf (t) = 12[(1+ 2)t3 + 2t2 (1 + 2)t + 2 2], f (t) = 12 [3(1+ 2)t2 + 2 2t 12],f (t) = 0t =

2 1 v t = 1+23 . Ta c b ng bi n thin

t

1

1+ 2

32

1 2

f (t) + 0 0 +

f (t)

1

386227 d

d d

0

2 + 2

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T b ng bi n thin ta cf (t) 2+ 2 ng th c x y ra khit = 2 hayP 2 2+ 2suy ramax P = 2 + 2 t c khit = 2 hayx = y =

22 .

N u2

t

1 th P 2 = 12 [(1

2)t 3

2t 2

(1

2)t + 2 + 2]. Xt hm s

g(t) = 12 [(12)t3 2t2 (1 2)t + 2 + 2], g (t) = 12 [3(12)t2 22t 1 + 2],g (t) = 0t = (2 + 1)(lo i) vt =

213 (lo i). Ta c b ng bi n thint 2 1g (t) +

g(t )

2

2

1

T b ng bi n thin suy rag(t) 1 ng th c x y ra khit = 1 hayP 2 1 suy ramin P = 1 t c khit = 1 hayx = 1, y = 0 ho cx = 0 , y = 1.

Nh n xt.Bi ton ny gi thi t v bi u th cP c cho d i d ng i xv i hai bi n nhng bi u th cP cha a c v d ngx + y,xy . V v y gi i bi ton ny, chng ta bnh phng bi u th cP . Sau y l m t bi ton tnt .

2.2.11 V d .Chox2 + y2 = 1 . Tm gi tr nh nh t, gi tr l n nh t c ath cP = x1+ y +

y1+ x .

Bi lm. tt = x + y, t gi thi tx2 + y2 = 1 hayxy = (x+ y)2

12 = t2

12 v b t ng th cxy

(x+ y)2

4 ta c2 t 2. Khi ta c

P 2 =(x + y)3 3xy (x + y) + ( x + y)2 2xy

1 + (x

+y

) +xy

+2xy

1 + ( x + y) + xy= t3 + 3 t + 2t2 + 2 t + 1

+2(t2 1)

|t + 1 |v it = 1.

N u1 < t 2 th P 2 = ( 2 1)t + 2 2. Do P 2 4 22 suy ramax P = 4 22 t c khit = 2 hayx = y =

22 .

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N u2 t < 1 th P 2 = (1 + 2)t + 2 + 2. Do P 2 4 + 22 suy ramin P = 4 + 22 t c khit = 2 hayx = y =

22 .

2.2.12 V d .Chox, y = 0 thay i tho mn(x + y)xy = x2

+ y2

xy . Tm gi trl n nh t c a bi u th cP = 1x 3 + 1y 3 .Bi lm. tx = 1a , y = 1b . Khi , gi thi t(x + y)xy = x2 + y2 xy tr thnh

a + b = a2 + b2 ab hay (a + b) = ( a + b)2 3ab v bi u th c

P = a3 + b3 = ( a + b)3 3ab(a + b) = ( a + b)[(a + b)2 3ab] = (a + b)2.

tt = a + b. p d ng b t ng th cab

(a + b) 2

4 , t gi thi t(a + b) = ( a + b)2

3ab

ta cab = t 2 t3 t2

4 hay t2 4t 0 suy ra0 t 4. Khi P = t2 16. V v ymax P = 16 t c khit = 4 haya = b = 2 hayx = y = 12 .

2.2.13 V d .Chox, y

R tho mnx2 + xy + y2 2. Tm gi tr l n nh t cbi u th cP = x2 xy + y2.

Bi lm. tt = x + y. T gi thi tx2 + xy + y2 2 hay (x + y)2 xy 2 ta cxy t

2

2. p d ng b t ng th cxy (x+ y)2

4 ta ct2

83 hay

83 t

83 .

Khi P = ( x + y)2 3xy 2t2 + 6 . B ng bi n thin hm sf (t) = 2t 2 + 6t 83 0 83f (t) + 0

f (t)

23

6 d

d d

23

T b ng bi n thin ta cmaxt[38 ;38 ]

f (t) = f (0) = 6 suy ramax P = 6 t c khi

(x ; y) = ( 2;2) ho c(x ; y) = ( 2;2).Nh n xt.Bi ton ny gi thi t l m t b t ng th c i x ng v bi

P c cho d i d ng i x ng v i hai bi n. gi i c bi ton n

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nh gi bi u th cP , i bi nt = x + y v tm i u ki n c a bi nt. Sau y l m ts bi ton v i nh h ng tng t .

2.2.14 V d .Chox, y > 0 tho mnx + y 1. Tm gi tr nh nh t c a bi u tP =

x2 + y2 + xyx + y

+1x

+1y

+3

xy.

Bi lm. tt = x + y. T gi thi t suy ra0 < t 1. p d ng b t ng th(x + y)2 4xy suy raxy t

2

4 . Khi P = (x+ y)2 xyx+ y +

x+ y+3xy 3t4 +

4( t+3)t 2 . Xt hm s

f (t) = 3t4 +4(t +3)

t 2 , f (t) = 34 4( t+6)

t 3 =3t 3 16t964t 3 , v i0 < t 1 ta c b ng bi n thin

t 0 1f (t) || f (t)

+ d

d d

674

T b ng bi n thin ta cmint(0;1]

f (t) = f (1) = 674 suy ramin P = 674 t c khix = y = 12 .

2.2.15 V d .Chox, y R tho mn2(x2 + y2) = xy + 1 . Tm gi tr nh nh tgi tr l n nh t c a bi u th cP = 7( x 4 + y4) + 4 x2y2.

Bi lm. tt = x2 + y2. T gi thi t2(x2 + y2) = xy + 1 ta cxy = 2 t 1 v pd ng b t ng th cx2 + y2 2xy v b t ng th cx2 + y2 2xy ta c25 t 23 .Khi P = 7( x2 + y2)210x2y2 = 33t2 +40 t 10. Xt hm sf (t) = 33t2 +40 t 10,f (t) = 66t + 40 , f (t) = 0t = 2033 . Ta c b ng bi n thin

t 25 2033 23f (t) + 0 f (t)

1825

609160

d d

d 2

19


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21/43


Xt hm s f (t) = t 2 t + 2 , f (t) = 2 t 1, f (t ) > 0 t [1;2]. V v ymax P =max

t[1;2]

f (t ) = f (2) = 4 t c khi(x ; y) = (0; 2) v min P = mint[1;2]

f (t) = f (1) = 2

t c khi(x ; y) = (0;

1).

Nh n xt.Bi ton ny gi thi t v bi u th cP c cho d i d ng khng x ng v i hai bi n. gi i c bi ton ny, chng ta nh gi gi tht = x2 + y2.

2.2.18 V d .Chox, y

R tho mn(x + y)3 + 4 xy 2. Tm gi tr nh nh t cbi u th cP = 3( x4 + y4 + x2y2) 2(x2 + y2) + 1 .

Bi lm. tt = x2 + y2. p d ng b t ng th c(x + y)2 4xy . T gi thi t(x + y)3 + 4 xy 2 suy ra(x + y)3 + ( x + y)2 2 hayx + y 1. Khi p d ng b t ng th cx2 + y2

(x+ y)2

2 ta ct 12 . p d ng b t ng th c(x2 + y2)2 4x2y2 tac P = 3[(x2 + y2)2 x2y2] 2(x2 + y2) + 1 3(x2 + y2)2

3(x 2 + y 2 ) 2

4 2(x2 + y2) + 1hayP 9t

2

4 2t + 1 . Xt hm sf (t) = 9t2

4 2t + 1 , f (t) = 9t2 2, f (t) = 0t = 49 .B ng bi n thin

t 12 +

f (t) +

f (t)

916

+

T b ng bi n thin ta cmint

[ 12 ;+ )f (t) = f ( 12) =

916 . V v ymin P = 916 t c khi

x = y = 12 .

Nh n xt.Bi ton ny gi thi t l b t ng th c i x ng v bi u P c cho d i d ng i x ng v i hai bi n. gi i c bi ton ny, chgi bi u th cP v tt = x2 + y2.

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2.2.19 V d .Chox, y > 0 tho mnx + y = 1 . Tm gi tr nh nh t c a bi u t

P =x

1 x+

y1 y

.

Bi lm. txy = t. p d ng b t ng th c(x+ y)2 4xy suy ra0 < t 14 . Khi P 2 = (x + y)

2

2xy xy (x+ y)1(x + y)+ xy+ 2xy1(x+ y)+ xy =

3t+1t + 2 t . Xt hm sf (t) = 3t+1t + 2 t,

f (t) = 1t 2 + 1t , f (t) = 0t = 1(lo i) vt = 0(lo i). B ng bi n thint 0 14

f (t) f (t)

+ d

d d 2

T b ng bi n thin ta cmint

(0; 14 ]f (t ) = f ( 14 ) = 2 suy ramin P = 2 t c khi

x = y = 12 .

Nh n xt.Bi ton ny gi thi t v bi u th cP c cho d i d ng i xv i hai bi n. Ngoi cch i bi nt = x + y ta c th gi i bi ton ny v i cch

bi nt = xy . Sau y l m t bi ton v i nh h ng tng t .2.2.20 V d .Cho cc s th cx, y tho mnx2 + y2 xy = 1 . Tm gi tr nhnh t, gi tr l n nh t c a bi u th cP = x

2 (x 2 +1)+ y 2 (y 2 +1)x 2 + y 2 3

.

Bi lm. T gi thi tx2 + y2 xy = 1 ta c 1 = x2 + y2 xy xy v 1 =x2 + y2 xy = ( x + y)2 3xy 3xy suy ra13 xy 1. tt = xy . Khi P = (x

2 + y 2 )2 2x2 y 2 + xy +1

(x 2 + y 2 )3= t

2 +3 t+2t2

. Xt hm s f (t) = t2 +3 t+2t2

,

f (t) = (t2)2

4(t2) 2 < 0t[13 ; 1].Do min P = min

t

[13 ;1]

f (t) = f (1) = 4 t c khix = y = 1 ho cx = y = 1vmax P = max

t[

1

3 ;1]f (t) = f (13) = 821 t c khi(x ; y) = (

33 ;

33 ) ho c(x ; y) =

(33 ;

33 ).

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2.2.21 V d .Chox, y > 0. Ch ng minh r ng(x4 + y 4 )

(x+ y)4 +xyx+ y 58 .

Bi lm.Do b t ng th c ng v ix, y th cng ng v itx,ty nn ta chu n

hox + y = 1 . tt = xy . Khi P = ( x2

+ y2)2

2x2y

2+ xy = [(x + y)

2

2xy ]2

2x2y2 + xy = 2 t4 4t2 + t + 1 . Xt hm s f (t) = 2 t4 4t2 + t + 1 v it(0; 12]. Tac f (t) = 8 t3 8t + 1 , dof (2)f (0) < 0, f (0)f ( 12 ) < 0, f ( 12)f (1) < 0 nn phngtrnh f (t) = 0 c ng 1 nghi mt0(0;

12 ) ta c b ng bi n thin

t 0 t0 12f (t) + 0 f

(t)

1

f (t0) d

d d 58


(0; 12 ]f (t) = f ( 12 ) =

58 t c khit = 12 v

x + y = 1 hayx + y = 1 , xy = 12 suy rax = y = 12 . Do P 58 ,x, y > 0.Nh n xt.Bi ton ny gi thi t v bi u th cP c cho d i d ng i x

v i hai bi n. gi i c bi ton ny chng ta dng phng php chu

2.2.22 V d .Chox, y > 0 tho mnx + y = 1 v m > 0 cho tr c. Tm gi trnh nh t c a bi u th cP = 1x 2 + y 2 + mxy .

Bi lm. tx = 12+ t, y = 12t v i0 |t| < 12 . Khi P = 24t 2 +1 4m4t 2 1 . Xt hm sf (t) = 24t 2 +1 4m4t 2 1 , f (t) =

16t(4 t 2 +1) 2 +

32mt(4t 2 1) 2

= 16t [4(m 1)t2 + m +1][4( m +1) t 2 + m 1](16 t 4 1)2

.N um 1 th f (t) = 0t = 0 . Kho min P = mint

(1

2 ;1

2 )f (t ) = f (0) = 2 + 4 m.

N u0 < m < 1 th f (t) = 0

t = 0 vt2 = 1m4(m +1)

v t2 = m +14(1m )

> 14(lo i) hay

t = 0 , t = 1m4(m +1) , t = 1m4(m +1) . Ta c b ng bi n thin

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24/43


t 12 1m4(m +1) 0 1m4(m +1) 12f (t) 0 + 0 0 +f (t)

+ d d

d 2m + 3m + 1

2 + 4m

d d d

2m + 3m + 1

+

T b ng bi n thin ta c

min P = mint(

1

2 ;1

2 )f (t) = f ( 1 m4(m + 1) ) = f ( 1 m4(m + 1) ) = 2 m + 3m + 1 .

Nh n xt.Bi ton ny gi thi t v bi u th cP c cho d i d ng i x

v i hai bi n. Ngoi nh ng nh h ng quen thu c bi ton c gi i php i bi n khc.

2.2.23 Bi t p.Chox, y 1. Ch ng minh r ng( 1x 1x 2 )( 1y 1y 2 ) 116 .H ng d n.B t ng th c cho tng ng v ix1x 2 y1y 2 116 hay

xy (x+ y)+1x 2 y 2 1

16 .

2.2.24 Bi t p.Chox, y = 0 . Tm gi tr nh nh t c a bi u th c

P =x4

y4+

y4

x4 2(x2

y2+

y2

x2) +

xy

+yx

.

H ng d n. tt = xy + yx , i u ki n|t | 2.2.2.25 Bi t p.Chox, y > 0 tho mnx2 + y2 = x 1 y2 + y1 x2. Tm gi trnh nh t c a bi u th cP = x2 + y2 + 1x 2 + 1y 2 .

H ng d n. tt = x2 + y2, v i0 < t 1.2.3. Bi ton tm gi tr nh nh t, gi tr l n nh t c a bi

th hi n tnh ng c p

Trong ph n ny chng ti trnh by m t s d ng bi ton tm gi tr ngi tr l n nh t c a bi u th c ch a hai bi n m gi thi t ho c bi u th c

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tnh ng c p. T xt hm s v tm gi tr nh nh t, gi tr l n nhs .

2.3.1 V d .Chox, y > 0 tho mnx2

+ y2

= 1 . Tm gi tr l n nh t c a bi u tP = y(x + y).

Bi lm. ty = tx t i u ki nx,y > 0 suy rat > 0. T gi thi tx2 + y2 = 1ta cx2 = 1t 2 +1 . Khi bi u th cP = x2t(t + 1) = t

2 + tt 2 +1 . Xt hm s f (t) = t

2 + tt 2 +1 ,

f (t) = t2 +2 t +1

(t 2 +1) 2 , f (t) = 0t = 2 + 1 v t = 1 2(lo i). Ta c b ng bi n thint 0 2 + 1 +

f (t ) + 0 f (t)

0

2+12

d d

d 1

T b ng bi n thin ta cmax P = maxt(0;+ )

f (t) = f (2 + 1) = 1+ 22 t c khix =

2122 v y =

2+122 .

Nh n xt.Bi ton ny gi thi t v bi u th cP c cho d i d ng ng cv i hai bi n. Sau y l m t s bi ton v i nh h ng tng t .

2.3.2 V d .Chox, y

R tho mnx2 + xy + y2 = 2 . Tm gi tr nh nh t, gi tl n nh t c a bi u th cP = x2 2xy + 3 y2.

Bi lm.N ux = 0 th y2 = 2 suy raP = 6 .N ux = 0 th ty = tx . Khi t gi thi tx2 + xy + y2 = 2 ta cx2 = 2

t2

+ t+1. V

v yP = 2(3t2

2t+1)t 2 + t +1 .Xt hm s f (t ) = 2(3t

2

2t+1)t 2 + t+1 , f (t ) =2(5t 2 +4 t3)(t 2 + t+1) 2 , f (t ) = 0 t =

2195 . Ta cb ng bi n thin

25


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t 219

5 2+19

5 + f (t) + 0 0 +

f (t)6

500+100 1975

d d d

5001001975

6


Rf (t) = f (2+ 195 ) = 500100

1975 t c khi

(x ; y) = (5 238+ 19 ; 238+ 19 (2 + 19)) ho c(x ; y) = ( 5 238+ 19 ; 238+ 19 (2 + 19)) vmax P = maxt

Rf (t) = f ( 2195 ) = 500+100

1975 t c khi

(x ; y) = (5 23819 ; 23819 (2 19)) ho c(x ; y) = ( 5 23819 ; 23819 (2 + 19)) .2.3.3 V d .Chox, y 0 tho mnx2 + y2 = 1 . Tm gi tr l n nh t c a bi u tP =

4x2 + 6 xy 52xy 2y2 1

.

Bi lm. N ux = 0 th t gi thi tx2 + y2 = 1 v y 0 suy ray = 1 . Khi P = 53 .N ux = 0 th ty = tx . T gi thi tx, y 0 vx2+ y2 = 1 suy rat 0 vx2 = 1t 2 +1 .Khi P = x

2 (4+6 t )5x 2 (2t2t 2 )1= 5t

2

6t+13t 2 2t+1. Xt hm s f (t ) = 5t 2 6t+13t 2 2t+1 , f (t) =

8t 2 +4 t4(3 t 2 2t+1) 2,

f (t) = 0t =12 v t = 1(lo i). Ta c b ng bi n thin

t 0 12 + f (t ) 0 +

f (t)

1 d

d d

1

53

T b ng bi n thin ta cf (t) < 53 t 0 v min P = mint[0;+ )

f (t) = f ( 12 ) = 1 t c khix = 25 v y =

15 . V v ymax P =53 t c khix = 0 , y = 1 v

min P = 1 t c khix = 25 v y = 15 .

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2.3.4 V d .Chox, y[2010;2011]. Tm gi tr nh nh t, gi tr l n nh t c

th cP = x+ yxy 2 (x2 + y2).

Bi lm. tt =xy . Khi P =

(t+1)( t 2 +1)t .

Xt hm s f (t) = (t +1)( t2 +1)

t , f (t) = 2 t + 1 1t 2 , f (t ) = 02t3 + t2 1 = 0.N u2010 x y 2011 th 20102011 t 1. Ta cf (t) > 0, t [20102011 ; 1] suy

ra max P = maxt[

2010

2011 ;1]f (t) = f (1) = 4 t c khix = y v min P = min

t[

2010

2011 ;1]f (t ) =

f ( 20102011 ) = 3 , 999005965 t c khix = 2010y2011 .N u2010 y x 2011 th 1 t 20112010 . Ta c f (t) > 0, t [1; 20112010 ]

suy ramax P = maxt[1;

2011

2010 ]f (t) = f ( 20112010 ) = 4 , 00099552 t c khix = 2011y2010 v

min P = mint

[1; 20112010 ]f (t) = f (1) = 4 t c khix = y.

Nh n xt.Bi ton ny bi u th cP c t v m u cng b c. gi i bi ny ta tt = xy v s d ng gi thi t xt cc tr ng h p x y ra c at.

2.3.5 V d .Chox, y

R tho mnx, y = 0 v x2 + y2 = x2y + xy 2. Tm gi trnh nh t, gi tr l n nh t c a bi u th cP = 2x + 1y .

Bi lm. ty = tx . T gi thi tx2 + y2 = x2y + xy 2 suy rax = t2

+1t 2 + t . Khi P = (t

2 + t )(2 t +1)t (t 2 +1) . Xt hm s f (t) = 2t

2 +3 t +1t 2 +1 , f (t) = 3t

2 +2 t +3(t 2 +1) 2 f (t) = 0 t =

1103 .B ng bi n thin

t 1103

1+ 103 + f (t ) 0 + 0

f (t)

2 d

d d

3102

3+ 102

d d

d 2

T b ng bi n thin ta cmax P = max f (t) = f ( 1+ 103 ) = 3+102 t c khi

(x ; y) = ( 20+2 1014+5 10 ;40+22 1042+15 10 ) v

min P = min f (t) = f ( 1103 ) = 3102 t c khi(x ; y) = ( 202

1014510

; 402210421510).

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2.3.6 V d .Chox, y > 0. Ch ng minh r ng 4xy 2(x+ x 2 +4 y 2 )3 18 .

Bi lm. tt = xy . T gi thi tx,y > 0 suy rat > 0. Khi b t ng th c

c n ch ng minh tng ng v i4t

(t+ t 2 +4) 3 18 hay t (t

2+ 4 t)

3

2. Xt hms f (t) = t(t2 + 4 t)3, f (t ) = ( t2 + 4 t )3 3t (t 2 +4 t )

3

t 2 +4 =(t 2 +4 t )

3 (t 2 +4 3t )t 2 +4 ,f (t) = 0t =

22 . Ta c b ng bi n thin

t 0 22 + f (t) + 0 f (t)

0

2 d

d d

0

T b ng bi n thin ta cmaxt

(0;+ )f (t) = f (22 ) = 2 hay t (t2 + 4 t)3 2 d u b ng

x u ra khit = 22 hayy = 2x.2.3.7 V d .Chox > y > 0. Ch ng minh r ngx+ y2 > xyln xln y > xy .

Bi lm.B t ng th c c n ch ng minh tng ng v i xy y > 0 suy rat > 1. Ch ng

minhln t > 2( t1)t+1 . Xt hm s f (t) =2( t1)t+1 ln t, f (t) = 4(t +1) 2 1t =

(t1)2

t (t+1) 2 0

t > 1. Ta c b ng bi n thint 1 +f (t) ||

f (t)

0 d

d d

T b ng bi n thin ta cf (t) < 0t > 1 hay

2(t1)t+1 < ln t t > 1.Ta ch ng minh xy 0.Xt hm s g(t) = t2 2tlnt 1, g (t) = 2( t lnt 1), g (t) = 2(1 1t ) 0, t > 0.

Suy ra pcm.

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2.3.8 V d .Chox, y > 0 tho mnxy y 1. Tm gi tr nh nh t c a bi u tP =

x2

y2+ 9

y3

x3.

Bi lm. tx = ty . T gi thi tx,y > 0 v xy y 1 ta c ty 2 + 1 = y hay1 = t2y + 1y 2t suy ra0 < t 12 . Khi P = t2 + 9t 3 . Xt hm s f (t) = t2 + 9t 3 ,f (t) = 2 t 27t 4 , f (t) = 0t =

5 272 (lo i). B ng bi n thint 0 12

f (t) || f (t)

+ d

d d

2894


(0; 12 ]f (t ) = f ( 12 ) =

2894 t c khi(x ; y) = (1; 2) .

Nh n xt.Bi ton ny gi thi t c cc s h ng khng cng b c. ton ny ta nh gi gi thi t rng bu c gi a hai bi nx, y .

2.3.9 Bi t p.Chox, y 0. Ch ng minh r ng3x3 + 7 y3 9xy 2.

B n c t gi i.

2.3.10 Bi t p.Chox, y 0. Ch ng minh r ngx4 + y4 x3y + xy 3.B n c t gi i.

2.3.11 Bi t p.Chox, y > 0. Ch ng minh r ng1x 3 + x3

y 3 + y3 1x + xy + y.

B n c t gi i.

2.3.12 Bi t p.Tm gi tr nh nh t c a bi u th cP = 3( x 2y 2 + y 2x 2 ) 8( xy + yx ) v ix, y = 0 .

H ng d n. tt = xy + yx , |t| 2.2.3.13 Bi t p.Cho0 < x < y < 1. Ch ng minh r ngx2 ln y y2 ln x > ln x lny .

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H ng d n.Xt hm s f (t) = ln t1+ t 2 .

2.3.14 Bi t p.Chox, y > 0. Tm gi tr nh nh t c a bi u th cP = x3 + y3 +7 xy (x+ y)

xy x 2 + y 2 .

B n c t gi i.

2.3.15 V d .Choa, b > 0 v x > y > 0. Ch ng minh r ng(a x + bx )y < (a y + by)x .

B n c t gi i.

2.4. Bi ton tm gi tr nh nh t, gi tr l n nh t c a bich a ba bi n

Trong ph n ny chng ti trnh by m t s d ng bi ton tm gi tr ngi tr l n nh t c a bi u th c ch a ba bi n b ng cch t n ph ho cqua m t bi n cn l i. T , chuy n c bi ton v bi ton tm gi trgi tr l n nh t c a hm s .

2.4.1 V d .Chox, y,z > 0 tho mnx + y + z = 1 . Ch ng minh r ng1xz + 1yz 16.Bi lm. tt = x+ y. T gi thi t ta cz = 1(x+ y) hayz = 1t v0 < t < 1. p

d ng b t ng th c(x + y)2 4xy hayxy t2

4 . Khi P = 1xz + 1yz = txy (1t ) 4

t 2 + t .Xt hm s f (t ) = 4t 2 + t , f (t) =

4(2t1)(t 2 + t )2, f (t) = 0t =

12 . Ta c b ng bi n thin

t 0 12 1f (t ) 0 +

f (t)

+ d

d d

16

+

T b ng bi n thin ta cmint(0;1)

f (t) = f ( 12 ) = 16 t c khix = y = 14 , z = 12 . Vv y 1xz + 1yz 16.

Nh n xt.Bi ton ny kh n gi n, ch c n th bi nz theo bi nx, y v ibi nt = x + y chng ta chuy n c bi ton v m t bi n.

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2.4.2 V d .Chox2 + y2 + z2 = 1 . Tm gi tr nh nh t, gi tr l n nh t cth cP = x + y + z + xy + yz + zx .

Bi lm. t t = x + y + z. p d ng b t ng th c Cauchy -Schwarz t(x + y + z)2 3(x2 + y2 + z2) suy ra3 t 3. Khi

P = ( x + y + z) +12

[(x + y + z)2 (x2 + y2 + z2)] =12

(t2 + 2 t 1).Xt hm s f (t ) = 12 (t 2 + 2 t 1), f (t) = 2 t + 2 , f (t) = 0t = 1. Ta c b ng bi nthin

t 3 1 3f (t)

0 +

f (t)

1 3 d d

d

1

1 + 3


[3;3]f (t ) = f (1) = 1 t c khit = 1

hay(x ; y; z) = ( 1;0;0) v cc hon v c a n;max P = max

t

[

3;3]

f (t) = f (3) = 1 + 3 t c khit = 3 hay (x ; y; z) =( 13 ;

13 ;13 ).

Nh n xt.Bi ton ny i x ng v i ba bi n, b ng cch tt = x + y + z chngta chuy n c v m t bi n. Sau y l m t s bi ton v i nh ht .

2.4.3 V d .Chox,y,z 0. Tm gi tr l n nh t c a bi u th cP =

1

x + y + z + 1 1

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

Bi lm. tt = x + y + z. p d ng b t ng th cxy + yz + zx (x + y+ z )2

3 vxyz

(x+ y+ z )3

27 . Ta c

P =1

(x + y + z) + 1 1

xyz + ( xy + yz + zx ) + ( x + y + z) + 1 1

t + 1 1

t 327 +

t 23 + t + 1

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=1

t + 1 27

(t + 3) 3.

Xt hm s f (t) = 1t+1 27(t +3) 3 , f (t) = 1(t +1) 2 + 81(t +3) 4 , f (t) = 0(t +3) 4 = [9( t +1)] 2

hay t = 0 v t = 3 . Ta c b ng bi n thint 0 3 +f (t) + 0

f (t)

0

18 d

d d

0

T b ng bi n thin ta cmaxt

[0;+

)f (t) = f (3) = 18 suy ramax P = 18 t c khi

x = y = z = 1 .

2.4.4 V d .Chox,y,z 0 tho mnx + y + z = 1 . Ch ng minh r ng

0 xy + yz + zx 2xyz 727

.

Bi lm. T gi thi tx,y,z 0, x + y + z = 1 ta c xy + yz + zx 2xyz =xy (1 z) + yz (1 x) + zx 0. D u b ng x y ra khi(x ; y; z) = (1; 0;0) v cc hon vc a n.

Do vai tr c ax,y,z bnh ng nn ta lun gi s cx = min {x,y,z }. T githi tx,y,z 0, x + y + z = 1 ta c0 x 13 v y + z = 1 x. p d ng b t ngth cyz

(y+ z ) 2

4 v1 2x > 0. Khi bi u th c

P = xy + yz + zx 2xyz = x(y + z) + yz (1 2x) x(1 x) + (1 2x)(y + z)2

4

= x(1

x) + (1

2x)

(1 x)24

=1

4(

2x3 + x2 + 1) .

Xt hm s f (x) = 2x3 + x2 + 1 v ix[0; 13 ]. Ta cf (x) = 6x2 + 2 x, f (x) = 0x = 0 v x = 13 . Ta c b ng bi n thin

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x 0 13f (x) +

f (x)

1

2827

T b ng bi n thin ta cmaxx

[0; 13 ]f (x) = f ( 13) =

2827 . Do max P = 14 maxx

[0; 13 ]

f (x) = 727

t c khix = y = z = 13 .

Nh n xt.Bi ton ny i x ng v i ba bi n, chuy n n v theo chng ta ph i ch n ph n t i di n, tm cch nh gi ph n t v bi

th c v theo ph n t i di n. Sau y l m t s bi ton v i nh ht .

2.4.5 V d .Chox,y,z 0 tho mnx + y + z = 1 . Ch ng minh r ng

x3 + y3 + z3 +154

xyz 14

.

Bi lm.Do vai tr c ax,y,z bnh ng nn ta lun gi s cx = min {x,y,z }.

T gi thi tx,y,z 0, x + y + z = 1 ta c0 x 13 v y + z = 1 x. p d ng b t ng th cyz

(y+ z )2

4 v 27x4 3 < 0. Khi bi u th c

P = x3 + y3 + z3 +154

xyz

= x3 + ( y + z)3 3yz (y + z) +15xyz

4= x3 + ( y + z)3 + yz [

15x4 3(y + z)]

= x3 + (1 x)3 + yz (27x

4 3) x3 + (1 x)3 +

(y + z)2

4(274

x 3)

=116(27x3 18x2 + 3 x + 4) .

Xt hm sf (x) = 116 (27x318x2+3 x+4) , f (x) = 116 (81x236x+3) , f (x) = 0x = 19v x = 13 . B ng bi n thin

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x 0 1913

f (x) + 0 f (x)

14

727

d

d d 14

T b ng bi n thin ta cmaxx[0;

1

3 ]f (x) = f (0) = f ( 13 ) =

14 . Do P 14 d u b ng x y

ra khi(x ; y; z) = ( 13 ; 13 ; 13) ho c(x ; y; z) = (0; 12 ; 12 ) v cc hon v c a n.

2.4.6 V d .Chox, y,z > 0 tho mnx + y + z = 1 . Tm gi tr nh nh t c a bith cP = x1

x +

y1

y +

z1

z.

Bi lm. Xt hm s f (x) = x1x, f (x) = 2x2(1x )1x

. Phng trnh ti p tuyc a th hm sy = f (x) t i i mM ( 13 ; 16 ) l y =

3122 (15x 1).

Ch ng minh x1x 3

122 (15x 1). N u0 < x < 115 th b t ng th c ng.N u 115 x < 1 th b t ng th c tng ng v i(3x 1)2(25x 1) 0 lun

ng, ng th c x y ra khix = 13.Tng t ta cy1

y

3122(15y 1), z1

z

3122 (15z 1).

C ng theo v ta cP 3122 [15(x + y + z) 3] = 62 . Do min P = 62 t ckhix = y = z = 13 .

Nh n xt.Bi ton ny i x ng v i ba bi n. Bi u th c ch a m i s trong m i s h ng ch ch a m t bi n. gi i bi ton ny chng ta th n phng php ti p tuy n. Sau y l m t bi ton v i nh h ng t

2.4.7 V d .Chox,y,z

(0;1) tho mnxy + yz + zx = 1 . Tm gi tr nh nhc a bi u th cP = x1x 2 +

y1y 2 +

z1z 2

.

Bi lm. Ta c P = x 2x (1x 2 ) +y2

y(1y 2 )+ z

2

z (1z 2 ). Xt hm s f (x) = 1x (1x 2 ) v i

0 < x < 1, f (x) = 3x2

1x 2 (1x 2 ) 2, f (x) = 0 x =

13 v x = 13(lo i). Ta c b ng bithin

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x 0 13 1f (x) 0 +

f (x)

+ d

d d 33

2

+

T b ng bi n thin ta c1x (1x 2 ) 33

2 x(0;1). V v y

P 33

2(x2 + y2 + z2)

332

(xy + yz + zx ) =33

2.

Do min P = 332 t c khix = y = z = 13 .

2.4.8 V d .Chox, y,z > 0 tho mnx2 + y2 + z2 = 1 . Tm gi tr nh nh t cbi u th cP = x 2y 2 + z 2 + y

2

z 2 + x 2 +z 2

x 2 + y 2 .

Bi lm.Ta cP = x2

1x 2 +y2

1y 2 +z 2

1z 2. Xt hm s f (x) = 1x (1x 2 ) v i0 < x < 1,

f (x) = 3x2

1x 2 (1x 2 )2, f (x) = 0x =

13 v x = 13(lo i). Ta c b ng bi n thinx 0 13 1

f (x) 0 +

f (x)

+ d

d d 33

2

+

T b ng bi n thin ta c1x (1x 2 ) 33

2 x(0;1).

V v yP 332 (x

3 + y3 + z3).p d ng b t ng th c AM-GM ta cx6 + x3y3 + x3y3 3x4y2; x6 + x3z3 + x3z3 3x4z2; ...;x3y3 + y3z3 + x3z3 3x2y2z2.

Khi 3(x6

+y6

+z6

+ 2x3y3

+ 2y3z3

+ 2x3z3

) x6

+y6

+z6

+ 3x4y2

+ 3x4z2

+ 3x2y4

+3y4z2 + 3 x2z4 + 3 y2z4 + 6 x2y2z2.

V v y ta ch ng minh c b t ng th c3(x3 + y3 + z3)2 (x2 + y2 + z2)3 hayx3 + y3 + z3 13 .Do P 3

32 (x

3 + y3 + z3) 32 suy ramin P = 32 t c khix = y = z = 13 .35


36/43


2.4.9 Bi t p.Chox,y,z 0 tho mn i u ki nx + y + z = 1 Ch ng minh

xy + yz + zx 27

+97

xyz.

H ng d n.N u79 x 1 th yz 97xyz v y + z 29. Suy raxy + yz < 29 < 27 .N u0 x < 79 th b t ng th c c n ch ng minh tr thnh(x + 1)(3 x 1)2 0.2.4.10 Bi t p.Choa,b,c l di ba c nh c a m t tam gic c chu vi bCh ng minh1327 a 2 + b2 + c2 + 4 abc < 12 .

B n c t gi i.

2.4.11 Bi t p.Chox, y,z > 0 tho mnx + y + z = 1 . Ch ng minh r ng2(x3 + y3 + z3) + 3( x2 + y2 + z2) + 12xyz

53

.

B n c t gi i.

2.4.12 Bi t p.Cho1 x,y,z 3 tho mnx + y + z = 6 . Ch ng minh r ng

x2 + y2 + z2 14.

H ng d n.Gi s x = max {x,y,z }. Khi 2 x 3 v P = x2 + y2 + z2 x2 + y2 + z2 + 2( y 1)(z 1). Xt hm s theo nx.2.4.13 Bi t p.Cho0 x,y,z 2 tho mnx + y + z = 3 . Ch ng minh r ng

x3 + y3 + z3 9.

H ng d n.Gi s x = max {x,y,z }. Khi 1 x 2 v P = x3 + y3 + z3 x3 + y3 + z3 + 3 yz (y + z). Xt hm s theo nx.2.4.14 Bi t p.Chox,y,z

R tho mnx2 + y2 + z2 = 2 . Tm gi tr nh nh t

gi tr l n nh t c a bi u th cP = x3 + y3 + z3 3xyz .H ng d n. tt = x + y + z.

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2.4.15 Bi t p.Chox,y,z 0 tho mnx + y + z = 3 . Tm gi tr l n nh t cbi u th cP = xy 2 + yz 2 + zx 2 xyz .

B n c t gi i.2.4.16 Bi t p.Chox,y,z 0 tho mnx + y + z = 3 . Tm gi tr l n nh t cbi u th cP = 9 xy + 10 xz + 22 yz .

H ng d n.P = 9 xy +10( x + y)z +12 yz = 10(x + y)2 +30( x + y12y2 +36 y3xy ).Xt hm s f (t ) = t2 + 3 t v it[0;3]. Suy ra

P = 10 f (x + y) + 12 f (y)

2xy

22 max

t[0;3]

f (t).

2.4.17 Bi t p.Chox, y,z > 0 tho mnx + y + z = 3 . Tm gi tr nh nh t cbi u th cP = 14+2 ln (1+ x )y +

14+2 ln (1+ y)z

+ 14+2 ln (1+ z )x.

H ng d n.p d ng b t ng th c Cauchy -Schwarz v xt hm s

f (t ) = 2 ln(1 + t) t.

2.4.18 Bi t p.Choa, b, c > 0 tho mn21ab + 2 bc + 8 ca 12. Tm gi tr nhnh t c a bi u th cP = 1a + 2b + 3c .

H ng d n. tx = 1a , y = 2b , z = 3c . Bi ton tr thnh: chox, y,z > 0 tho mn2x + 4 y + 7 z 2xyz . Tm gi tr nh nh t c aP = x + y + z. p d ngz 2x +4 y2xy 7 tac

P

x + y +

2x + 4 y 14x + 14xx(2y

7x )

x + y +2

x

+2x + 14x

2xy 7= x +

11

2x+

2xy 72x

+2x + 14x

2xy 7.

p d ng b t ng th c AG-MG v xt hm sf (x) = x + 112x + 2 1 + 7x 2 .2.4.19 Bi t p.Chox,y,z

R tho mnx2 + y2 + z2 = 9 . Ch ng minh r ng

2(x + y + z) xyz + 10 .37


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H ng d n.p d ng b t ng th c Cauchy-Schwarz ta c

[2(x + y + z) xyz ]2 = [x (2 yz ) + 2( y + z)]2 [x2 + ( y + z)2][(2yz )2 + 2 2].

hay[2(x + y + z) xyz ]2 (9 + 2yz )(y2z2 4yz + 8) . tt = yz ta c[2(x + y + z) xyz ]2 2t3 + t2 20t + 72 .

Gi s |x| |y| |z| t gi thi tx2 + y2 + z2 = 9 suy ra|x|2 3. V v y

|t| = |yz | y2 + z2

2=

9 x22 3.

2.4.20 Bi t p.Chox,y,z[0;1]. Ch ng minh2(x

3 + y3+ z3)(x2y+ y2z + z2x) 3.H ng d n.Xt hm s g(x) = f (x,y,z ) v i nx vy, z l tham s . Hm sg(x)

t gi tr l n nh t t ix = 0 ho cx = 1 nhngf (0) f (1) nn ta ch ng minhf (1) 3. Ti p t c xt hm sh (y)...2.4.21 Bi t p.Choa, b, c > 0. Ch ng minh r ng

(2a + b + c)2

2a2 + ( b + c)2+

(2b + c + a )2

2b2 + ( c + a )2+

(2c + a + b)2

2c2 + ( a + b)2 8.H ng d n.Do b t ng th c ng v ia,b,c th cng ng v ita, tb, tc . V v y

gi s a + b + c = 3 . p d ng phng php ti p tuy n.

2.4.22 Bi t p.Chox,y,z[1006; 2012]. Tm gi tr l n nh t c a bi u th c

P =x3 + y3 + z3

xyz.

H ng d n.Gi s 1006 x y z 2012. ty = kx , z = tx , khi 1 k t 2 v P = 1+ k

3 + t 3kt .

Ch ng minhP 1+ k 3 +2 3

2k . Xt hm s f (k) =k 3 +9

2k v ik[1;2].2.4.23 Bi t p.Cho cc s th ca,b,c khng ng th i b ng0 tho mna2+ b2+ c2 =2(ab + bc + ca ). Tm gi tr nh nh t v gi tr l n nh t c a bi u th c

P =a3 + b3 + c3

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

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H ng d n. tx = 4aa + b+ c , y = 4ba + b+ c , z = 4ca + b+ c . Khi ta cx + y + z = 4 vxy + yz + zx = 4 . p d ng b t ng th c(y + z)2 4yz suy ra0 x 83 .

Khi P = 132 (x3 + y3 + z3) = 132 (3x3

12x2 + 12x + 16) .

2.4.24 Bi t p.Chox,y,z[1;2]. Tm gi tr l n nh t c a bi u th c

P = ( x + y + z)(1x

+1y

+1z

).

H ng d n.Gi s 1 x y z 2 suy ra(1xy )(1yz ) 0 v(1yx )(1zy ) 0.Nhn ra r i c ng theo v ta c

P = (x

y+

y

x) + (

y

z+

z

y) + (

x

z+

z

x) + 3

5 + 2(

x

z+

z

x).

tt = xz , v it[12 ; 1]. Ta c(2 t)( 12 t) 0 hay t + 1t 52 .

2.4.25 Bi t p.Chox + y = 0 . Tm gi tr nh nh t c a cc bi u th c

P = x2 + y2 + (1 xyx + y

)2 v Q = 4 x2 + 13 y2 + ( xy 12x + y

)2.

H ng d n. tz = 1xyx+ y .

2.4.26 Bi t p.Chox,y,z 0 tho mnx2 + y2 + z2 = 1 . Tm gi tr l n nh t cbi u th cP = 6( y + z x) + 27xyz .

H ng d n.P 6[ 2(y2 + z2) x] + 27x.y 2 + z 2

2 = 6[ 2(1 x2) x] +27x (1x

2 )2 .

2.4.27 Bi t p.Choa, b, c > 0. Ch ng minh r nga 3(a + b)3 + b3

(b+ c) 3 +c3

(c+ a )3 38 .H ng d n. tx = ba , y = cb, z = ac suy raxyz = 1 . B t ng th c tr thnh

1(1 + x)3

+ 1(1 + y)3

+ 1(1 + z)3 38 .

p d ng b t ng th c AM-GM ta c1(1+ x )3 + 1(1+ x )3 + 18 32(1+ x )2 .Ta c n ch ng minh b t ng th c1(1+ x )2 + 1(1+ y) 2 + 1(1+ z )2 34 .p d ng b t ng th c1(1+ x )2 + 1(1+ y)2 11+ xy x, y > 0.

39


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Ta cP = 1(1+ x )2 + 1(1+ y)2 + 1(1+ z )2 11+ xy + 1(1+ z )2 = z1+ z + 1(1+ z )2 .Xt hm s f (z) = z1+ z + 1(1+ z )2 .

2.4.28 Bi t p.Chox, y,z > 0 tho mnx2

+ y2

+ z2

= 1 . Ch ng minh r ngx

y2 + z2+

yz2 + x2

+z

x2 + y2 33

2.

B n c t gi i.

2.4.29 Bi t p.Chox,y,z[0;1] tho mnx + y + z = 1 . Tm gi tr nh nh t v

gi tr l n nh t c a bi u th cP = 1x 2 +1 + 1y 2 +1 + 1z 2 +1 .

H ng d n.p d ng phng php ti p tuy n v b t ng th c1

x2 + 1+

1y2 + 1 1 +

1(x + y)2 + 1

x, y > 0; x + y 1.

2.4.30 Bi t p.Chox,y,z 0 tho mnx + y + z = 1 . Tm gi tr nh nh t vgi tr l n nh t c a bi u th cP = 1x1+ x + 1y1+ y + 1z1+ z .

H ng d n.p d ng b t ng th c

1x1+ x 1 x v b t ng th c

1 x1 + x + 1 y1 + y 1 + 1 (x + y)1 + ( x + y)x + y 45 .2.4.31 Bi t p.Choa, b, c > 0. Ch ng minh r ng 2aa + b + 2bb+ c + 2cc+ a 3.

H ng d n. tx = ab , y = bc , z =

ca suy raxyz = 1 v b t ng th c

11 + x2

+1

1 + y2 2

1 + xyxy 1.

2.4.32 Bi t p.Chox, y,z > 0 tho mn(x + y + z)3 = 32xyz . Ch ng minh r ng383 1655

2 x4 + y4 + z4

(x + y + z)4 9

128.

H ng d n.Gi s x + y + z = 4 . tt = xy + yz + zx.

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2.4.33 Bi t p.Chox, y,z > 0 tho mnx + y + z 1. Ch ng minh r ng

3(x + y + z) + 2(1x

+1y

+1z

) 21.

H ng d n. tt = x + y + z.

2.4.34 Bi t p.Chox, y,z > 0 tho mnx2 + y2 + z2 = 1 . Ch ng minh r ng

(1x

+1y

+1z

) (x + y + z) 23.

H ng d n. tt = x + y + z.

2.4.35 Bi t p.Chox, y,z > 0. Tm gi tr nh nh t c a bi u th c

P =2(x + y + z)3 + 9 xyz

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

H ng d n.Gi s x + y + z = 1 v z = min {x,y,z }suy ra0 < z 13 .2.4.36 Bi t p.Cho xy + yz + zxx 2 + y 2 + z 2 = 17. Tm gi tr nh nh t, gi tr l n nh t c ath cP = x

4 + y 4 + z 4

(x + y+ z )4 .

H ng d n.Gi s x + y + z = 1 v z = min{

x,y,z

}suy ra0 < z

1

3.

2.4.37 Bi t p.Choa, b, c, d > 0 tho mna 2 + b2 = 1 v c d = 3 . Ch ng minhr ngP = ac + bd cd 9+6

24 .

H ng d n.p d ng b t ng th c Cauchy-Schwarz ta c

P (a 2 + b2)(c2 + d2) cd = 2d2 + 6 d + 9 d2 3d.

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K T LU N

Bi vi t thu c cc k t qu sau.1. Trnh by c th cch tm gi tr nh nh t, gi tr l n nh t c a hm2. H th ng m t s d ng bi ton tm gi tr nh nh t, gi tr l n nh

bi u th c ch a hai bi n b ng cch th m t bi n qua bi n cn l i.3. H th ng m t s d ng bi ton tm gi tr nh nh t, gi tr l n

m t bi u th c ch a hai bi n b ng cch t n ph theo tnh i x ngt = x + y,t = x2 + y2 ho ct = xy .

4. H th ng m t s d ng bi ton tm gi tr nh nh t, gi tr l n nhbi u th c ch a hai bi n b ng cch t n ph theo tnh ng c pt = xy .

5. H th ng m t s d ng bi ton tm gi tr nh nh t, gi tr l n nhbi u th c ch a ba bi n b ng cch t n ph ho c th hai bi n qua m t

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TI LI U THAM KH O

[1] Ban t ch c k thi (2001 n 2010),Tuy n t p thi Olympic 30 thng 4, Nhxu t b n gio d c.

[2] PGS.TS Nguy n Qu Dy, Th.S Nguy n Vn Nho, TS V Vn ThoTuy n t p 200 bi thi v ch ton , Nh xu t b n gio d c.

[3] Ban bin t p (2001 n 2010),T p ch Ton h c v Tu i tr , Nh xu t b ngio d c.

[4] Ph m Kim Hng (2008),Sng t o b t ng th c, Nh xu t b n gio d c.

[5]www.math.vn , 2011.

[6]www.violet.vn , 2011.

[7]www.mathscope.org , 2011.

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