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.
 A: 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 $-1\lt x\lt0$,
$$
\log(1+x)\lt\frac{2x}{2+x}
$$
A: Proof by contradiction:
$$\ln(1+x) \le \frac{2x}{2+x}$$
$$\ln(1+x) \le 2 - \frac{4}{2+x}$$
$$\ln(1+x) - 2 < -\frac{4}{2+\ln(1+x)}$$
$$\ln^2(1+x) < 0$$
A: Hint: Let $f(x)=\ln(1+x)-\frac{2x}{2+x}$, and show that $f(0)=0$, and that $f'(x)>0$ for all $x>0$.
A: 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 $x \geq 4$.
A: At zero both are zero. Their derivatives are
$$\frac 1 {1+x},\qquad \frac 4 {(2+x)^2}=\frac 1 {(1+x/2)^2}$$
A: Also for sufficiently large values of $x$ it is quite obvious since
\begin{equation}
\frac{2x}{2+x} = 2 - \frac{4}{2 + x}
\end{equation}
and therefore $\frac{2x}{2+x}$ has upper boundary while $\ln(x+1)$ is unbounded.
A: We change the question to find an upperbound:$$\ln(1+x) > \frac{nx}{n+x}$$
Let's do the following substitution: $\left( 1+x \right) ={ e }^{ h }.$ The question becomes, proving $$h>\frac { n\left( { e }^{ h }-1 \right)  }{ n+{ e }^{ h }-1 } ,$$ where $h>0.$ 
We search for $n$: $$n<\frac { h{ e }^{ h }-h }{ { e }^{ h }-h-1 } $$. For larger values of $h$ we know that the equation will satisfy therefore we have to prove for smaller values of $h$. $$\lim _{ h\rightarrow o }{ \left( \frac { h{ e }^{ h }-h }{ { e }^{ h }-h-1 }  \right)  } =2.$$ Therefore the upper bound of $n$ is $2$. 
