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.

To show that the inequality holds for $x \geq 0$ note that
$$ \log(1+x) = \int_1^{1+x} \frac{1}{t}dt. $$
Approximate the function $t \mapsto \frac{1}{t}$ by its tangent at $t = 1 + \frac{x}{2}$ to get
$$ \frac{1}{t} \geq -\frac{1}{(1+\frac{x}{2})^2}(t - 1 - \frac{x}{2}) + \frac{1}{1 + \frac{x}{2}} = -\frac{4 t}{(x + 2)^2} + \frac{4}{x+2}$$
for all $t \geq 1$.  In particular
$$ \log(1+x) \geq \int_1^{1+x} \left(-\frac{4 t}{(x + 2)^2} + \frac{4}{x+2}\right) dt = \frac{2x}{x+2}.$$
This lower bound is optimal among all those that can be obtained by using such a tangent approximation.  In general it depends what you mean by "optimal".  For example
$$ \log(1+x) - \frac{2x}{a x + 2} = \frac{a-1}{2} x^2 + O(x^3) $$
which implies that $a \geq 1$ to get a possible lower bound.  But for $a > 1$ this lower bound is worse than for $a=1$.  On the other hand
$$ \log(1+x) \geq \frac{8x}{3(x+4)} $$
for all $x\geq 0$ and this is a better lower bound for $x \geq 4$.
