x y 属于r xy=1 x大于y 求（x的平方加上y的平方）除以（x-y）的最小值

x,y∈R,xy=1,y＜x
∴(x²+y²)/(x-y)
=[(x-y)²+2xy]/(x-y)
=[(x-y)+[2/(x-y)]
∵x＞y,得x-y＞0
∴可以用均值不等式,得
[(x-y)+[2/(x-y)]
≥2[√(x-y)]{√[2/(x-y)}
=2√2
当且仅当x-y=2/(x-y),即x-y=√2时取得等号
即最小值是2√2