Use Jensen's inequality to show $\frac{2x}{2+x} < \log(1+x) < \frac{2x+x^2}{2+2x}$ for $x>0$

Question

Use Jensen's inequality to show $\frac{2x}{2+x} < \log(1+x) < \frac{2x+x^2}{2+2x}$ for $x>0$.

I can show this without Jensen's inequality, but I'd like to see what that form of the proof looks like.

Without Jensen's, start with the inequality $\log(1+x) < x$ for $x>0$, integrate both sides to arrive at the upper bound. Also, $\log(1+x) > 1-\frac{1}{x+1}$ for $x>0$ (by substituting $x=1/u-1$ into the previous inequality). Integrate both sides to arrive at the lower bound.

Answer 1

Hermite-Hadamard inequality :

if a function $ƒ : [a, b] → R$ is convex, then the following chain of inequalities hold: $$f\Big(\frac{a+b}{2}\Big)\leq \frac{1}{b-a}\int_{a}^{b}f(x)dx\leq \frac{f(a)+f(b)}{2}$$

Now take $f(x)=\frac{1}{x}$ and $a=1$ and $b=x+1$ to get a lower and a upper bound . Without this refinement I think it's hopeless .