Question

1.已知x>y>0,且xy=1,则(x^2+y^2)/(x-y)的最小值是

Answer 1

(x^2+y^2)/(x-y)
=[(x-y)^2+2xy]/(x-y)
=[(x-y)^2+2]/(x-y)
=(x-y)+[2/(x-y)]
>=2根号[(x-y)*2/(x-y)]=2根号2
{当且仅当x-y=根号2时等号成立}