Question

不等式 ：已知x>y>0 xy=1 求证(x^2+y^2)/(x-y)≥2√2

Answer 1

证明：

原式=[（x-y)^2+2xy]/(x-y)
=[（x-y)^2+2]/(x-y)

=(x-y)+2/(x-y)
因为x＞y＞0 ∴x-y＞0

所以(x-y)+2/(x-y) >=2√[（x-y)*2/(x-y)]=2√2

Answer 2

首先(x²+y²)/(x-y) = (x-y)²/(x-y)+2xy/(x-y) = (x-y)+2/(x-y) (∵xy = 1).
由x > y, 有x-y > 0, 2/(x-y) > 0.
由均值不等式(x-y)+2/(x-y) ≥ 2√((x-y)·2/(x-y)) = 2√2.