Prove that for $\displaystyle -1<x\leq0$
$\displaystyle log_{e}(1+x)\geq\frac{tan^{-1}x}{1+x}$
Define: f(x)=ln(1+x) and g(x)=1/{1+x^2} and use Mean value theorem - Wikipedia, the free encyclopedia on$\displaystyle [x,0] \subset (-1,0]$ where $\displaystyle x>-1$.
Not sure... trying it...
Edit:
It's works!
Read my post again(read the Cauchy mean value theorem).
$\displaystyle c\in (-1,0], [x,0]\subset (-1,0] x>-1 $
$\displaystyle \frac{f(x)-f(0)}{g(x)-g(0)}=\frac{\ln(1+x)-0}{\arctan(x)-0}=\frac{f'(c)}{g'(c)}=\frac{\frac{1}{c+1}}{\frac{ 1}{1+c^2}}=\frac{1+c^2}{c+1}>\frac{1}{1+x}$
Prove the last inequality.
Here is one way to do it. Multiply by 1+x (which doesn't change the inequality since 1+x>0). Then we want to show that
$\displaystyle f(x) \overset{\text{d{e}f}}{=} (1+x)\ln(1+x) - \tan^{-1}x \geqslant 0\quad(-1<x\leqslant0).$
Since f(0) = 0 it will be sufficient to show that f(x) is decreasing in that interval. So we want to show that its derivative is negative. But
$\displaystyle f'(x) = \ln(1+x) + 1 - \frac1{1+x^2}.$
We want to show that $\displaystyle f'(x)\leqslant0$ for $\displaystyle -1<x\leqslant0$. To do that, repeat the previous argument. Since f'(0)=0 it is sufficient to show that f'(x) is increasing in that interval. So we want to show that its derivative is positive. But
$\displaystyle f''(x) = \frac1{1+x} + \frac{2x}{(1+x^2)^2} = \frac{(1+x)^2 + 3x^2+x^4}{(1+x)(1+x^2)^2},$
which is clearly positive for x>–1.