①
f(1) = x + k/x = 1 + k = 2
k = 1
f(x) = x + 1/x
设x1>x2>1
f(x1) - f(x2)
= (x1 - x2) + (1/x1 - 1/x2)
= (x1 - x2)·(x1·x2 - 1)/(x1·x2)
x1 > x2
x1·x2 > 1
所以
f(x1) - f(x2) > 0
f(x1) > f(x2)
在（1,+∞）上的单调递增
②
f(x) = x + k/x
设x1>x2>1
f(x1) - f(x2)
= (x1 - x2) + (k/x1 - k/x2)
= (x1 - x2)·(x1·x2 - k)/(x1·x2)
x1 - x2 > 0
x1·x2 > 1 >0
(1)
当k ≤ 1
x1·x2 - k > 0, f(x1) > f(x2), f(x)在（1,+∞）上的单调递增
(2)
当k > 1
当x1, x2 > √k
x1·x2 - k > 0, f(x1) > f(x2), f(x)在（√k,+∞）上的单调递增
当x1, x2 < √k
x1·x2 - k < 0, f(x1) < f(x2), f(x)在（1, √k）上的单调递减