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/(1-x) +b^2
所以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/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