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

用分析法。

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a^2/x=(1-x+x)a^2/x
=(1-x)a^2/x + a^2

b^2/(1-x)=(1-x+x)b^2/(1-x)
=xb^2/祥迟巧(1-x) +b^2

所旦行以a^2/x+b^2/(1-x)
=(1-x)a^2/x+xb^2/(1-x)+a^2+b^2
>=2√[(1-x)a^2/谨键x*xb^2/(1-x)]+a^2+b^2
=2ab+a^2+b^2

所以a^2/x+b^2/(1-x)>=(a+b)^2

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