(b + 3c)^ (a + 3c)- (Ạ 2013) Giai: Giai:
Nhan xet: Bleu thuc A c6 dang ciing bac va 2 bien a,b dol xung.
Dat a = xc; b = yc => x; y > 0 => (x + l)(y + l) = 4c:>x + y + xy = 3. Do rinh doi xung cua x, y ta dat x + y = s;xy = p=>s + p = 3. xung cua x, y ta dat x + y = s;xy = p=>s + p = 3.
Mat khac ta c6: > 4p 3 < s + — => 2 < s < 3
Ta CO danh gia sau: Voi m,n > 0 thi m'^ + n-' = (m + n)'^ - 3mn(nr m'^ + n-' = (m + n)'^ - 3mn(nr
Ap dung vao bai toan ta c6:
m^ + n^ = (m + n)^ -3mn(m + n) > (m + nf - | ( m + n)^ (m + n) = ^ ( m + n)^ A > 8 A > 8 <=> A > 8 a b + - b + 3c a + 3c - 8 (x + y)^ +3(x + y ) - 2 x y c n3 n3 y+ 3 x+3 -7(x + y)^-2xy 3(x + y) + xy + 9 Thay x + y = s,xy = s- 3 ta c6 A >(s-1)'' -7s^ + 2 s-6 = £(s)
Xet ham so f(s) = (s -1)^ - Vs^ + 2 s- 6 voi 2 < s < 3 ta c6
s + 1
f(s) = 3( s- i r -
Vi s> 2=>3(s-l)^ >3. s + 1
Vs^ +2s-(
^ f(s + 1) 9
nen f'(s) > 0. Tu do ta c6 f(s) > f(2) = 1 - V2 . Dau bang xay ra khi va chi khi a = b = c a = b = c
Vi dvi 17: Cho cac so thyc duang x , y , z 6 [ l ; 4 ] , x > y ; x > z . Tim GTNN cua bieu thuc sau: P = -—^^-^ + ^ + (Ạ 2011) bieu thuc sau: P = -—^^-^ + ^ + (Ạ 2011)
2x+3y y + z z+x Giai: Giai: Ta thay bieu thuc P la bieu thuc dang cap Ta viétlai P = ^ - ^ i — + —1_ +
2x + 3y 1 + ^ 1 + ^ De y rang —.— = — > 1. De y rang —.— = — > 1.
y z y
la CO bo de: Voi a,b > 0,ab > 1 thi : + > — ^ (*)
1 + â 1 + b^ 1 + ab
That vay (*) tuang duang voi: ^ ) (^^ ^) > 0 bat dSng thuc
( l + a 2 j ( l + b 2 j ( l + ab)
nay la hien nhien diing.
Ap dyng vao bai toan ta c6: P ^
X y y -I D a t t = / - > l d o x , y e r i ; 4 l = > t < 2 t^ 2 Ta c 6 P > — + — = f(t) 2t^+3 1 + t t^ 2
Xet ham so f(t) = — + z—- voi t e
f ( t ) = -2
2t2+3 1 + t t^(4t-3) + 3t(2t-l) + 9 t^(4t-3) + 3t(2t-l) + 9
1; 2 ta CO
Dau bang xay ra khi va chi khi
<0 suyra f(t)>£(2) = — 33 33 t = 2 x^^ z^o x = 4;y = l;z = 2 z y IX; f, . (adsbygoogle = window.adsbygoogle || []).push({});
C A C Vf D U S A U s e M I N H H Q A C H O BAT D A N G THCTC 3
B i^ N K H 6 N G D6\G NHLTNG C 6 T H ^ Q U Y V £ M6T
B i^ N N H O C A C B A T D A N G THLTC T R U N G G I A N V A K H A I T H A C Y ^ U T 6 D A N G C A P T H A C Y ^ U T 6 D A N G C A P
Vi dv 18: Cho cac so thuc khong am a,b,c thoa man: â + = c(c - l),c> 0 . Tim gia tri nho nhat va GTLN cua P = â+b^+c^ Tim gia tri nho nhat va GTLN cua P = â+b^+c^
(a + b + c) GiAi: GiAi:
GTNN:
Ta thay voi moi x,y € R thi - x y + y^ = ^(x + yf+^{x- yf > i ( x + y)^
Theo gia thiet ta c6: a + b + c = c
Tir do ta suy ra: a + b^ < 1 ^ 0 < ^ < 1 c c Ta c6: P = â+b^+c^ a + b V + 2 (a + b + c) (a + b + c) ^ 0;1 2 / u >
Khao sat ham so f(t) =
2(t + l)'
voi tefO;!] taco f'(t) = (t + 1)' (t + 1)'
<OVterO;l
suy ra ham so f(t) nghjch bien . Do do ta c6: f(t) > f(l) = -3 3 3 3 Khi a = b = l;c = 2 thi P = - . Vay Pmin = - .
8 o
GTLN:!
Ta CO : (a + b + c)^ = â + b^ + + 2(ab + be + ca) > â + b^ + Suy ra P S1 khi a = b = 0;c = 1 thi P = 1 v|iy Pmax = 1 Suy ra P S1 khi a = b = 0;c = 1 thi P = 1 v|iy Pmax = 1
Vi du 19: Cho cac sóthuc a,b,c e [1,2].
Tim gia tri nho nhát ciia bieu thuc P = • (a + b)2 + 4(ab + be + ca) + 4(ab + be + ca) Giai:
Ap dung bat d5ng thue Cosi bp hai so dang 4ab < (a + b)2, ta d l dang nhan thay thay P> (a + bf e2+(a + b)2+4e(a + b) P>- a b — + — 1 + fa b^ 2 a — + — + 4 + — S Dat - + - = t , t 6 [ l , 4 ] c c
Ta se khao sat ham so: P = • t^
t^ + 4 t + l • = f(t) voi t6ri;4 Xethamso f(t) = Xethamso f(t) =
t^ + 4 t + l te 1;4 ta eo f'(t) = 4t^+2t
f t 2 + 4 t + l
2" > 0 voi mpi
t e [l;4] nen minP = - dat dugc khi t = 1 tuc a = b = l,c = 2
6 ' Vi du 20: Cho eae so thuc duong a, b, e thoa man: Vi du 20: Cho eae so thuc duong a, b, e thoa man:
â + b^+ê+ab-2bc-2ea-0. Tim gia trj nho nhat eiia P = Tim gia trj nho nhat eiia P =
(a + b - c ) 2 â + b^ a + b Giai: Giai:
Gia thiet bai toan eo the viet lai thanh: (a + b - e)^ = ab nen ta dv doan dau bang xay ra khi a = b = c . Tir do ta nghi den viee dua ve bien t = — bang xay ra khi a = b = c . Tir do ta nghi den viee dua ve bien t = — (adsbygoogle = window.adsbygoogle || []).push({});
C 2
a2+b2 daụ Vi vay ta nghT den each dung theo bien t - ^ ^ daụ Vi vay ta nghT den each dung theo bien t - ^ ^
.1 a + b
Taco P = - + - 2 £ ^ = Q
Bay gia ta d e y :
K h i dau bang xay ra t h i = ^ I ^ ^ ' = ^17^ 2 c2 2ab nen ta phan tich:
2 a b^ a 2 + b 2 2ab (a + b)^ (a + b)^ p a b ( a + b)2 (a + b)^ a + b 2ab 2c Dat t = — d o (a + b - c) 2= a b< f c ^ a + b 4 1- a + b 1 1 c ^ 3 <-<=>-< < - 4 2 a + b 2
Xet ham so f(t) = + 1 vol t e [1;3] ta c6:
f ( t ) = 2t + l > 0 = ^ m i n f ( t ) = f ( l ) = 2khi t = l o a = b = c
V i d v 21: Cho cac so thuc diromg a,b,c thoa man: 3b^c^ +â =2(a + bc). 4 4
T i m gia tri nho nhat cua bieu thuc: P = — + —j + Z ^
^ • be (a + b r (a + c r G i a i
Gia thiet c6 the viet lai n h u sau: 3b^c2 - 2bc = 1 - (a - 1 ) < 1 be < 1
Ta can tan d u n g dieu kien nay de danh gia P
Ta c6: (« + b) 2 " ( a 2+ b c ) ' T u o n g t u c 1 ( a ^ + b c K b + c) (3 + ^) (â+bc) T u do: P > â + 4 ( a ^ + b c K b + c) b = a 2+ - i - > a 2 + a + b c (â+bcKb + c) (â+bc)(b + c ) j -> 4
Xet ham so f(a) = â + — tren [0;+oo) a ^ + l
„ , ^ 2 a( a 2+ 3 ) ( a - l ) ( a + l ) ^
T a c o f (a) = —^ - = Ooa = l
(a2+l)2
Lap bang bien thien ham so f(a) tren [0;+<») ta thay £(a) > f ( l ) = 3 N h u v a y m i n f ( a ) = 3 o a = 1 m i n P = 3ci>a = b = c = l
a ^ + l
256
i dv 22: Cho x, y, z la cac só thi^c khong am thoa man: xy + yz + zx > 0 T i m gia tri nho nhát cua bleu thiic: T i m gia tri nho nhát cua bleu thiic:
^ ^ r + 4Vz + l +2^x + 2y + 4 ' } ' P ^
7x(y + z) + 2z2 M^ + 4z)
GiAị P=i
7x(y + z) + 2z^ Vy(x + 4z) + 47z+T + 27 x + 2y+ 4
Taco: P = 2(V4z + 4 +^x + 2 y + 4) + - • = 2Q + R 7x(y + z) + 2z2 Vy(x + 4z)
• Xet: Q = V4z + 4+7x + 2y + 4 > 2 + ^x + 2y + 4z + 4 > 2 + 74 + x + 2 y + 3z
B po Va + 4 + Vb + 4 > 2 + V4 + a + b(Va;b > 0)dau bang xay ra k h i va chi k h i (adsbygoogle = window.adsbygoogle || []).push({});
B ab = 0 )
• x e t R = ^ ^ Vx(y + z) + 2z2 Mx + 4z) 4^(xy+ xz + 2z2)(yx + 4zy)
2^/2 4V2
72xy + xz + 2 z 2 + 4 z y V('^ + 2z)(z + 2y) x + 2y + 3z" .4>/2
Dat t = x + 2y + 3 z > 0 . S u y ra: P>2(2 + >/4 + t )
Xet ham so f(t) = 2(2 + TiTt) + i ^ , t e (0; +00) 1 4V2 1 4V2
t
= 0 <::> t = 4 £(t) ^ f(4) = 4 + 5^2 .
l + t t
Vay m i n P = 4 + 5V2 <=>x = 2;y = l ; z = 0
V i dv 23: Cho cac so thvc d u o n g x,y,z thoa man dieu kỉn x^ + y^ + = 3 .
I T i m gia trj nho nhat cua bieu thuc P = 1 — + + . ^
( y + z)^ (z + x f (x + y)^
G i a i :
2
Su d u n g bat dang thuc (a + b) < 2
x y z
â+b^
V y ^ + z ^ z^+x^ x^ + y^ 1,3-x^ 3 - y 2 3 - ; ^ + - y - , _ £ .
= i [ x( 3- x= ) % ( 3 ' - y ' ) \ ( 3- . ^ ) J
Vi X , y , z duang va x^ + + = 3 nen x e ^0;^/3 j .
Xetham f(x) = x ( 3- x 2 ) tren (0;>/3).
Taco f'(x) = -3x^ + 3 , f ' ( x ) - 0 o x = l va f (x)>0<=> x e(0;l). Suyra f ( x ) < f ( l ) = 2 vai mpi x€(0;S). Suyra f ( x ) < f ( l ) = 2 vai mpi x€(0;S).
Do do
Tuong tu ta cung c6 - JL > 2 _ va-
, ( 3 - y ^ ) 3 - 2 '
1 x^ + v^ + z^ 3
X = y = z = 1.
Vay gia trj nho nhat ciia P la -^^ dat khi x = y = z = 1.
Chii y: De chung minh bat dang thuc — -r- > — ta cung c6 the su dung (adsbygoogle = window.adsbygoogle || []).push({});
x l 3- x ^ j 2
bat d3ng thuc Cosi nhu sau: Ta CO Ta CO x(3-x2)) . x 2 . ( 3 - x 2 ) ( 3 - x 2 ) 1 o . . 2 / , _ 2 U , _ v 2 U i j a.2x^.(3-x^)(3-x^) f 2 x 2 + 3 - x 2 + 3 - x 2 = 4 x^ • Suy ra x ( 3- x 2) < 2 . D o do ^^^_^2 j " T " . . .
V i dv24: Cho x,y,z>0,x>z.rim gia trj Ion nhat cua
258 Ta CO P = 1 1 Ta CO P = 1 1 Giii: 1 1 + 1 + ^ ^2 Z y> 1 + ^ V z X Dat a = —,b = —,c = - . X y z Khido a , b , c > 0 , c > l va P= , ^ + ^ Vi abc = l , c > l = > a b < l . Ta chung minh: , ^ + . ^ < ^ Vl + â Vl + b^ Vl + ab
That vay theo bát dang thuc Bunhiacopxki ta c6:
1
V i T ^ ^ V i T b ^
Ta can chung minh:
1 1
U + â ^ l + b ^ ;
V l + â l + b^y Vl + ab
1 1 ^ 2 ( a - b r ( l - a b ) •>0 l + â l + b^ 1 + ab (l + âXl + b^Kl + ab) l + â l + b^ 1 + ab (l + âXl + b^Kl + ab)
Dieu nay la luon diing voi mpi ab < 1. Dang thuc xay ra khi a = b .
Quay tro l^i bai toan ta c6: P < , ^ + • ^ Vl + ab Vl + c Vl + ab Vl + c
V
1 ^ 1 2 V c + l
c Khao sat ham só f(c) = ^^^^^ tren [ l ; +00) ta c6: Khao sat ham só f(c) = ^^^^^ tren [ l ; +00) ta c6:
ViT7
f(c) = - 2-V^ £'(c) = 0 o c - 4 2(l + c)7c(c + l ) 2(l + c)7c(c + l )
Lap bang bieh thien ta thu du^c:
^^^^ Maxf(c) - Vs khi c = 4 hay x = 4z 2y
dy 25: Cho x > y > z > 0. Tim GTNN cua bieu thuc: P = ^ + P = ^ +
Giii: / \2 X Taco p = ^ ^ + — ^ + - 1 - ^ 1 - ^ 8 V Z X Dat a = ^ , b = —,c = - . ^ - 1 a b K h i d o a,b,c>0;abc = l ; 0 < a , b < l , c > l va P = + +
Ap dyng bát dang thiic Co si ta c6: > ab 2>/ab 1-a 1 - b 8 ( > / ^ - l ) 4N/ab 1- a 1 - b i{1-a){l-h) 2- ( a + b) Vi a + b > 2 N / a b = i > 2 - ( a + b ) < 2 - 2 N / a b va 2 - ( a + b ) > 0 > J; D o d o P . - ^ . 2 - + • 1-%/^ 8 ( 7 ^ - l ) yfc-l 8 (N/ ^ - I ) 8 (V ^ - I )
Xet ham so g(c) = ^ . tren (l;+oo) thi
8(vc - I j
3 c ' - 4 c V ? - 1 6
Lap bang bien thien cvia ham g(c) tren (l;+Qo) suy ra P > 4, d4ng thuc xay
ra khi va chi khi a = b = abc = l,c = 4 <=> a = b = -^,0 = 4 < » X = 2y = 4z.
Vay minP = 4, dat dup-c khi x = 2y = 4z. (adsbygoogle = window.adsbygoogle || []).push({});
g <=>c = 4
V i dv 26: Cho cac so thi^c duong x, y, z sao cho x^ + y^ + z^ = 3 . Tim GTNN
cuabieuthiic P = , + , Giai Ap dyng bat dang thuc Co si ta c6:
^ x ^ + x y V y ^ + x y ^ ( x 2 + x y ) ( y 2 + x y ) ( x ^ + x y ) + ( y 2 + x y ) V ? 7 y
T u d o t a c o : P> , ^ . ^ = - ^ ^ 2 ^ ^
Xet ham so f(z) = ^ J = = + ^ tren (O; l ) ta c6
f(z) = 2z 2V3 2z(l + z f- 2 7 3 (I - Z ^ ) N / I ^
( i. z) ^ ( i- z^ ) v r
f'(z) = 0 o 2 z( l + z f - 273(1 - z^ j V l T ^ = 0 » 4z3 - 8z2 + 9z - 3
» ( 2 z- l) ( 2 z 2 - 4 z + 3J = 0 o z = ^ . Lap bang bien thien ciia ham = / ^ + ^ tren (O; l ) ta c6 P > dang thuc xay ra khi va chi
1 \/3 1
khi x = y,z = - , x ^ + y ^ + z ^ = l < » x = y = - y - , z = -^.
Vay minP = —— ,datdugckhi x = y = — , z = - .
3 2 2
V i d^ 27. Cho cac so thyc duong a, b, c.
2 2 2
Tim gia trj nho nhát cua bieu thuc P = — - — - + — - — - +
{a + bf {h + cf 3(c + a)' Giii: 4 2 2 3 1 + i i, Ta c6: P =
I Dat x = - , y = - , z = - . Khi do x,y,z la cac so thuc duong thoa man
I a b c " • 1 1 4 I xyz = 1 va P = + + (1 + x f (1 + y f 3( l + z f 1 1 1 Tachungminh - + -> (1) (1 + x f ( l + y f 1 + xy
Th?it vay, BDT (1) <=> xy (x - yf + ( l - xy)^ > 0, l u o n dung.
D a n g thiic xay ra k h i va chi k h i x = y = 1.
1 4
T u BDT (1) va xyz = 1 ta c6: P > + -i + >^y 3 ( 1 + z ) 3 1 + z z 4 - + 3 ( 1 . z f ; .
Xet h a m so f (z) = — + ^—r tren (O; +co).
, . ^ 4 ( z - l ) ( z + 3) , ,
T a c o f'(z) = ^ ^ = ^ >L—L^í(z) = 0<=>z = l ( d o z > 0 ) . ( z . l f ( l . z r ( l . z f
2
Lap bang bien thien ciia ham f(z) tren (0;+oo) ta suy ra P ^ - / dang thuc
o (adsbygoogle = window.adsbygoogle || []).push({});
xay ra k h i X = y = 1
z = l c:>x = y = z = l hay a = b = c . Vay m i n P = - d?it duoc k h i a = b = c
3
V i du 28: Xet cac so thuc a > b > c > 0 thoa man a + b + c = 1 .
a / b 24
T i m gia t r i nho nhat ciia bieu thuc. P = . I +. / + -
b + c Vc + a sVSa + 5b
Giai:
Truoc het ta chung m i n h BDT : , / - ^ + J-^ > 2. / ^'^ \) V b + c V c + a V a + b + 2 c V o i cac so thuc d u o n g thoa man a + b > 2c.
That vay, ta chung m i n h voi cac so a, b, x, y, m, n t h i : (â + b^) (x^ + y^ j f m ^ + n^) > (axm + b y n f (*) A p d u n g BDT Cauchy ta c6: 3axm x3 m^ • + — r- + - âfb^ x3+y3 m3 + n3"|77b^)(777) 3byn m^ + n^ b-^ y^ n â+b^ x3 + y3 m^ + n^ a/fâ + b^)(x3 + y ^ j p + n^) Cgng theo v e h a i BDT tren ta dugc BDT can chung m i n h : A p d u n g (*) ta c6: