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

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.

• This is the first step in a proof I'm following to show Stirling's approximation. – MONODA43 Jun 3 '20 at 0:46

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 .