- Néu hai só x,y trai dáu ta suyra xy <
Khaosathamso f(z )= —(1 z)^+ z^ tren [0;l] suyra
= 108 - 36(xy + yz + zx) + 27xyz = 108 - 36x(y + z) + yz(27x - 36) A
N2
+ 27xyz
Vi 27x-36<0,yz:S y + z suy ra
A > 108 - 36x(3 - x) + fc^i(27x - 36) = 108 - 36x(3 - x) + i^-A(27x - 36) Mat khac: 108 - 36x(3- x) + ^^'^^ (27x - 36) > 27 <:> x(x -1)^ > 0. Mat khac: 108 - 36x(3- x) + ^^'^^ (27x - 36) > 27 <:> x(x -1)^ > 0.
Suy ra dieu phai chung minh. Dáu bang xay rakhi x = y = z = l
3 3 (y + z ) ' ( 3 - x ) ' Cach 2: Ta su dung bat dang thuc trung gian: y + z > — = -^^— Cach 2: Ta su dung bat dang thuc trung gian: y + z > — = -^^—
Trong 3 so x,y,z luon ton tai hai so 6 ciing mpt phia so voi 1.
Khong mat tinh tong quat ta gia su hai só do la y,z suy ra
( y - l ) ( z - l ) > 0 < = > y z > y + z - l = 2 - x . Do do 15xyz> 15x(2-x) suy ra 4^x^ + y^ + z^ j + 15xyz > 4x^ + (3 - x)^ + 15x(2 - x). Bay gio ta chi can chung 4^x^ + y^ + z^ j + 15xyz > 4x^ + (3 - x)^ + 15x(2 - x). Bay gio ta chi can chung
minh: 4x3 ^ ^3 _ ^^3 ^ ^5^(2 - x) > 27 o - 6x2 + 3x > 0 » 3^^^ -1)2 > 0.
Nhung bát dang thuc nay luon diing. Dáu bang xay ra khi x = y = z = 1
Nhan xet: Vai vỉc phat hi^n: x = y = z = 1 thi dáu bang cua bát dSng thuc xay rạ Ta c6 dieu hien nhien sau: (x - l)(y - l)(y - l)(z - l)(z - l)(x -1) > 0 xay rạ Ta c6 dieu hien nhien sau: (x - l)(y - l)(y - l)(z - l)(z - l)(x -1) > 0 Vi vay mpt trong ba tich: (x - l)(y - l);(y - l)(z - l);(z - l)(x -1) phai c6 mpt
só khong am.
Gia su (y - l)(z -1) > 0 ta cung suy ra dupe: y z > y + z - l = 2 - x
V i d\f. 32: Cho cac só thuc x, y, z e [-1,2] thoa man x + y + z = 1.
Tim gia trj Ion nhát va GTNN ciia bieu thuc: A = x^ + ý* + z^ Giai: Giai:
GTLN:
Cty TNHH MTV DWH kh,u, , ,
t'.:.j 1
Ta tháy r^ng trong ba só x,y,z phai c6 it nhát hai so cung dáụ Gia sir hai sódo la x, y suy ra xy > 0. sódo la x, y suy ra xy > 0.
Khi do dey rang, x^ +ý* = (x + y)'' - xy(4x2 + 6xy + 4 y 2 ) < (x + y)"* =(l -z)"*.
Xet ham f(z) = ( l - z ) ' * + z \ z G [ - l , 2 ] . , . ,
De dang nhan tháy t'(z) = 0 chi c6 nghiem duy nhát z = ^ .
B Tir do suy ra Maxf(z) = Max|f(-l),f(l),f(2)| = f(-l) = f(2) = 17
P Do dáu dang thuc c6 the xay ra khi (x;y;z) = (-l;2;0)ho|c (2;-l;0) nen ta
ket luan dupe maxA = 17 GTNN: GTNN:
-St
- Néu hai só x, y trai dáu ta suy ra xy < 0
x'' + ý' = (x + y)'* - xy(4x2 + 6 x y + 4 y 2 ) >(x + y)'* = (1 - z)^
Khaosathamso f ( z ) = ( l - z ) ^ + z ' * suy ra min A = —<=> z - —;xy = 0 (adsbygoogle = window.adsbygoogle || []).push({});
16 2
- Néu ba s o ' x , y , z cung dáu thi x , y , Z G 0;1 • ' Ta CO x'*+ý* > —( x ^+ y 2 ] > l ( x + y)'* Ta CO x'*+ý* > —( x ^+ y 2 ] > l ( x + y)'*
2 V ' 8
Khao sat ham so f(z) = —(1 - z)^ + z^ tren [ 0 ; l ] suy ra 8 8
^ fmin = —<=>x = y = z = i . Vay MinA = —
I 27 3 27