已知x>y>o,且xy=1,若x^2+y^2>= a(x-y)恒成立,求实数a的取值范围
问题描述:
已知x>y>o,且xy=1,若x^2+y^2>= a(x-y)恒成立,求实数a的取值范围
答
因为x>y>o,
所以x-y>0,2/(x-y)>0
x^2+y^2
=x^2+y^2-2xy+2xy
=(x-y)^2+2
=(x-y)[(x-y)+2/(x-y)]
>=(x-y) 2根号2
所以a