设 1 > x1 > x2 > 0
f(x1) - f(x2)
x1 - x2 + 2 / x1 - 2 / x2
= x1 - x2 + 2(x2 - x1) / (x1 * x2)
= (x1 - x2)(1 - 2 / (x1 * x2)) < 0
f(x1) < f(x2)
所以为减函数