x2+64/y(x-y)
=（x-y+y)^2+64/y(x-y)
=(x-y)^2+y^2+2y(x-y)+64/y(x-y)
因为(x-y)^2+y^2>=2y(x-y)
所以原式>=4y(x-y)+64/y(x-y)>=2根号(4*64)=2*2*8=32
当且仅当：x-y=y,4y(x-y)=64/y(x-y)时取等号,此时y=2,x=4