f（x）=lnx x+1 +1 x
所以f(x)-lnx x-1 =1 1-x2 (2lnx-x2-1 x )
考虑函数h(x)=2lnx-x2-1 x (x＞0),
则h′(x)=2 x -2x2-(x2-1) x2 =-(x-1)2 x2
所以当x≠1时,h′（x）＜0而h（1）=0,
当x∈（0,1）时,h（x）＞0可得1 1-x2 h(x)＞0；
当x∈(1,+∞)时,h(x)＜0,可得1 1-x2 h(x)＞0
从而当x＞0且x≠1时,
f(x)-lnx x-1 ＞0即f(x)＞lnx x-1