1 |
1+x |
1 |
(1+x)2 |
x |
(1+x)2 |
当-1<x<0时,f′(x)<0,f(x)在(-1,0)上单调递减,
当x=0时,f′(x)=0,
当x>1时,f′(x)>0,f(x)在(1,+∞)上单调递增,
所以x=1是f(x)的极小值点也是最小值点,
所以f(x)的极小值=f(0)=0;
(2)由(1),f(x)≥f(0)=0,从而ln(1+x)≥
x |
1+x |
要证lna-lnb≥1-
b |
a |
a |
b |
b |
a |
令1+x=
a |
b |
x |
1+x |
1 |
x+1 |
b |
a |
a |
b |
b |
a |
x |
1+x |
b |
a |
1 |
1+x |
1 |
(1+x)2 |
x |
(1+x)2 |
x |
1+x |
b |
a |
a |
b |
b |
a |
a |
b |
x |
1+x |
1 |
x+1 |
b |
a |
a |
b |
b |
a |