Proving: $\frac{2x}{2+x}\le \ln (1+x)\le \frac{x}{2}\frac{2+x}{1+x}, \quad x>0.$ $$\begin{equation}\frac{2x}{2+x}\le \ln (1+x)\le \frac{x}{2}\frac{2+x}{1+x}, \quad x>0\end{equation}$$
I found this inequality in this paper: http://ajmaa.org/RGMIA/papers/v7n2/pade.pdf (Equation (3)). 
How exactly can I prove it? I tried induction but to no avail...

Let $f(x) = \frac{2x}{x+2}$, $g(x) = \ln(1+x)$, $h(x) = \frac{x^2 + 2x}{2x+2}$. 
$f(0) = 0$, $g(0) = 0$, $h(0) = 0$.
$f' = \frac{2(x+2)-2x}{(x+2)^2} = \frac{4}{(x+2)^2}$, $g' = \frac{1}{1+x}$, $h'= \frac{(2x+2)(2x+2)-2(x^2+2x)}{(2x+2)^2}=\frac{2x^2+4x+4}{(2x+2)^2}$
Now it is easy to check that
$$f'=\frac{4}{(x+2)^2} \leq g' = \frac{1}{1+x} \leq h'=\frac{2x^2+4x+4}{(2x+2)^2}$$
