Let t= 1/x, t>- 1 and t is not equal to 0.

Only t & gtln( 1+t) needs to be proved.

Prove:

Constructor f(t)=t-ln( 1+t)

f '(t)= 1-( 1/( 1+t))

There are two situations:

(1) when t >; 0， 1+t & gt； 1，0 & lt； 1/( 1+t)& lt； 1, so there is always f' (t) >; 0

F(t) in t >; 0 is monotonically increasing, so f (t) >; F(0)=0 holds.

T & gtLn( 1+t) hold.

(2) When-1

0 & lt 1+t & lt； 1

1(1+t) >1,so f' (t) < 0.

f(x)in- 1； F(0)=0 holds.

T & gtLn( 1+t) hold.

Synthesis (1)(2) proves that when t>- 1 and t is not equal to 0.

T & gtLn( 1+t) hold.

This also proves that when x∈(-∞,-1)∩(0, +∞), 1/x >; Ln( 1+ 1/x) is maintained.

Method 2

Constructor f(t)=t-ln( 1+t)

f'(t)

= 1-( 1/ 1+t)

Let the derivative f'(t)=0.

1-( 1/ 1+t)=0

Find the unique extreme point t=0.

After testing, t=0 is the minimum point of function f(t).

So f(t)>=f

f(t)>=f(0)=0-ln 1=0

So t-ln (1+t) >: 0 is a constant, that is, t & gtLn( 1+t).