(x2+y2)/(x-y)= (x2+y2-2xy+2xy)/(x-y) 因为xy=1,所以 =[(x-y)^2+2]/(x-y) =(x-y)+2/(x-y) 因为x>y>0所以(x-y)>0 所以有不等式的定理知道 （x-y)+2/(x-y)>=2根号下[（x-y)*2/(x-y)]=2根号2 而此时（x-y)^2=2 符合上...