已知x>0,y>0,x+y=1,求(1+x^-2)(1+y^-2)的最小值＝?此时x=?y=?

Answer 1

x>0,y>0

x+y=1
(x+y)^2=1
x^2+y^2=1-2xy

(x-y)2≥0
x^2+y^2 ≥ 2xy

1-2xy ≥ 2xy
xy ≤ 1/4

(1+x^-2)(1+y^-2)
= 1+x^-2+y^-2+(xy)^(-2)
= 1+1/x^2+1/y^2+1/(x^2y^2)
= 1 + (x^2+y^2+1)/(x^2y^2)
= 1 + (1-2xy+1)/(x^2y^2)
= 1 + (2-2xy)/(x^2y^2)
= 1 + 2{[1/(xy)^2-1/(xy)}
= 1 + 2{[1/(xy)^2-1/(xy)}
= 1 + 2{[1/(xy)^2-1/(xy)+1/4-1/4}
= 1 + 2{1/(xy)-1/2}^2 - 1/2
= 1/2 + 2{1/(xy)-1/2}^2
xy≤1/4
1/(xy)≥4
{1/(xy)-1/2}^2 ≥49/4
1/2 + 2{1/(xy)-1/2}^2 ≥ 1/2 + 49/2 = 25
(1+x^-2)(1+y^-2)的最小值＝25，x=y=1/2

追问

怎么想到的啊？？做这类题目有什么技巧吗？？？

追答

条件中又x，y
题目中有x^2，y^2
∴要把x+y=1两边平方
又：题目中能够化出xy
∴要用到基本不等式x^2+y^2≥2xy

Answer 2

对了