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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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1.3.4 Bi t p.Tm gi tr nh nh t v gi tr l n nh t c a hm s
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.
1.3.7 Bi t p.Tm gi tr nh nh t v gi tr l n nh t c a hm s
f (x) = sin 2 x x, v ix[2
;2
].
B n c t gi i.
1.3.8 Bi t p.Tm gi tr nh nh t v gi tr l n nh t c a hm s
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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Gio vin: Tr n nh Hi n - Tr ng THPT ng Thc H a - Ngh An
t 1 2f (t) || f (t)
+ d
d d 4 + 32
T b ng bi n thin ta cmin P = mint
(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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Gio vin: Tr n nh Hi n - Tr ng THPT ng Thc H a - Ngh An
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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Gio vin: Tr n nh Hi n - Tr ng THPT ng Thc H a - Ngh An
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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Gio vin: Tr n nh Hi n - Tr ng THPT ng Thc H a - Ngh An
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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Gio vin: Tr n nh Hi n - Tr ng THPT ng Thc H a - Ngh An
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
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Gio vin: Tr n nh Hi n - Tr ng THPT ng Thc H a - Ngh An
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
T b ng bi n thin ta cmin P = mint
(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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Gio vin: Tr n nh Hi n - Tr ng THPT ng Thc H a - Ngh An
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
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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
T b ng bi n thin ta cmin P = mint
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
T b ng bi n thin ta cmin P = mint
(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
T b ng bi n thin ta cmin P = mint
[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
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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.
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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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Gio vin: Tr n nh Hi n - Tr ng THPT ng Thc H a - Ngh An
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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