Why does $ \frac{2x}{2+x}$ provide a particularly tight lower bound for $\ln(1+x)$ for small positive values of $x$? EDIT:
My question was poorly worded. 
I wasn't asking about showing $\ln(1+x) > \frac{2x}{2+x}$ for $x>0$. 
What I wanted to know is why the lower bound provided by $ \frac{2x}{2+x}$ was so tight for small positive values of $x$. This is addressed in robjohn's answer.

As I mentioned in a comment, the power series near $x=0$ are
$$
\begin{align}
\log(1+x)&=x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4)\\
\frac{2x}{2+x}&=x-\frac{x^2}{2}+\frac{x^3}{4}+O(x^4)\\
\log(1+x)-\frac{2x}{2+x}&=\frac{x^3}{12}+O(x^4)
\end{align}
$$
So near $x=0$, the value, and the first and second derivatives match. That means the functions $\log(1+x)$ and $\frac{2x}{2+x}$ match to second order. Not quite as simple, $\frac{x(6+x)}{6+4x}$ matches $\log(1+x)$ to third order. Rational approximations to functions are called Padé Approximations.
Furthermore,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\left(\log(1+x)-\frac{2x}{2+x}\right)
&=\frac{\mathrm{d}}{\mathrm{d}x}\left(\log(1+x)+\frac{4}{2+x}-2\right)\\
&=\frac1{1+x}-\frac4{(2+x)^2}\\
&=\frac{x^2}{(1+x)(2+x)^2}
\end{align}
$$
So, for $x\gt-1$, $\log(1+x)-\frac{2x}{2+x}$ is an increasing function. At $x=0$, $\log(1+x)-\frac{2x}{2+x}=0$.
Therefore, for $x\gt0$,
$$
\log(1+x)\gt\frac{2x}{2+x}
$$
and for $-1\lt x\lt0$,
$$
\log(1+x)\lt\frac{2x}{2+x}
$$
