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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.

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

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  • $\begingroup$ Thanks! Looks like you can use Jensen's to prove the lower bound in the Hermine-Hadamard inequality. $\endgroup$ – MONODA43 Jun 3 '20 at 18:48

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