已知x>y>0,xy=1,求证(x^2+y^2)/(x-y)≥2根号2

已知x>y>0,xy=1
(x^2+y^2)/(x-y)
=(x^2-2+y^2+2)/(x-y)

‍=(x^2-2xy+y^2+2)‍‍‍‍‍‍‍‍‍‍/(x-y)‍


=[(x-y)‍‍‍‍^2+2]‍‍/(x-y)=x-y+2/(x-y)‍≥‍2‍根号2