Why does $ \frac{2x}{2+x}$ provide a particularly tight lower bound for $\ln(1+x)$ for small positive values of $x$?

Hint: Let $f(x)=\ln(1+x)-\frac{2x}{2+x}$, and show that $f(0)=0$, and that $f'(x)>0$ for all $x>0$.

Proof by contradiction: $$\ln(1+x) \le \frac{2x}{2+x}$$ $$\ln(1+x) \le 2 - \frac{4}{2+x}$$ $$\ln(1+x) - 2 < -\frac{4}{2+\ln(1+x)}$$ $$\ln^2(1+x) < 0$$

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 $-1\lt x\lt0$, $$ \log(1+x)\lt\frac{2x}{2+x} $$