Question

高二数学题已知a>0b>0,0<x<1,求证(a^2/x)+(b^2/(1-x))大于等于（a+b)^2

a^2表示a的平方
谢谢

Answer 1

倒推法
假设:(a^2/x)+(b^2/(1-x))大于等于（a+b)^2
==>a^2-a^2x+b^2x >=a^2x-a^2x^2+b^2x^2-b^2x^2+2abx_2abx^2
==>2a^2x-a^2x^2-b^2x^2+2abx-2abx^2-a^2<=0
==>(a+b)^2x^2-2a(a+b)x+a^2>=0
==>[(a+b)x-a]^2>=0
==>[a(x-1)+bx]^2>=0
a>0b>0,0<x<1
得证