<s>[INST] Proving $x+\sin x-2\ln{(1+x)}\geqslant0$ 
Question: Let $x>-1$, show that
  $$x+\sin x-2\ln{(1+x)}\geqslant 0.$$

This is true. See http://www.wolframalpha.com/input/?i=x%2Bsinx-2ln%281%2Bx%29
My try: For
$$f(x)=x+\sin x-2\ln{(1+x)},\\
f'(x)=1+\cos{x}-\dfrac{2}{1+x}=\dfrac{x-1}{1+x}+\cos{x}=0\Longrightarrow\cos{x}=\dfrac{1-x}{1+x}.$$
So

$$\sin x=\pm\sqrt{1-{\cos^2{x}}}=\pm \dfrac{2\sqrt{x}}{1+x}$$

If $\sin x=+\dfrac{2\sqrt{x}}{1+x}$, I can prove it. But if $\sin x=-\dfrac{2\sqrt{x}}{1+x}$, I cannot. See also http://www.wolframalpha.com/input/?i=%28x-1%29%2F%28x%2B1%29%2Bcosx
This inequality seems nice, but it is not easy to prove.
Thank you.
 [/INST] Let $f(x)=x+\sin{x}-2\ln(1+x).$
Thus, $$f'(x)=1+\cos{x}-\frac{2}{1+x}=2-\frac{2}{1+x}-(1-\cos{x})=\frac{2x}{1+x}-2\sin^2\frac{x}{2}\leq0$$
for all $-1<x\leq0,$ which says that for $-1<x\leq0$ we have
$$f(x)\geq f(0)=0.$$
Let $x\geq5.$
Thus, since $$\left(x-2\ln(1+x)\right)'=1-\frac{2}{1+x}=\frac{x-1}{x+1}>0,$$ we obtain:
$$f(x)\geq x-1-2\ln(1+x)>4-2\ln6>0.$$
Id est, it's enough to prove our inequality for $0\leq x\leq5.$
We see that $f'(x)=0,$ when $\sin\frac{x}{2}=\sqrt{\frac{x}{x+1}}$ or $\sin\frac{x}{2}=-\sqrt{\frac{x}{x+1}}.$
Also,  $\sin\frac{x}{2}+\sqrt{\frac{x}{x+1}}\geq\sqrt{\frac{x}{x+1}}\geq0$ for all $0\leq x\leq5$.
Thus, all critical points of $f$ on $[0.5]$ gives the following equation.  $$\sin\frac{x}{2}=\sqrt{\frac{x}{x+1}}.$$ 
We see that $\sin\frac{x}{2}$ decreases on $[\pi,5]$ and $\sqrt{\frac{x}{x+1}}$ increases on $[\pi,5]$.
Also, we have  $$\sin\frac{\pi}{2}>\sqrt{\frac{\pi}{\pi+1}}$$ and $$\sin\frac{5}{2}<\sqrt{\frac{5}{5+1}},$$ 
which says that the equation $$\sin\frac{x}{2}=\sqrt{\frac{x}{x+1}}$$ has an unique root $x_1\in[\pi,5].$
Also, we see that $f'(x)<0$ for all $x\in\left(\pi,x_1\right)$ and $f'(x)>0$ for all $x\in(x_1,5),$ which says $x_1$ gives a minimum point and  $x_{min}=4.06268...$.
Now, we'll prove that the equation 
$$\sin\frac{x}{2}=\sqrt{\frac{x}{x+1}}$$ has exactly two roots on $[0,\pi]$.
Indeed, on this set it's equivalent to
$$\sin^2\frac{x}{2}=\frac{x}{x+1}$$ or
$$\frac{1}{\sin^2\frac{x}{2}}=1+\frac{1}{x}$$ or $\tan\frac{x}{2}=\sqrt{x}$. 
But $\tan$ is a convex function and $\sqrt{x}$ is a concave function, which says that the equation $\tan\frac{x}{2}=\sqrt{x}$ has two roots maximum.
$0$ is one of them and $$\sqrt{\frac{\pi}{2}}-\tan\frac{\pi}{4}>0$$ and $$\sqrt{\frac{3\pi}{4}}-\tan\frac{3\pi}{8}<0,$$
which says that our equation has last root $x_2\in\left(\frac{\pi}{2},\frac{3\pi}{4}\right)$ and $x_2=1.88176...$.
Also, we  saw that $f'(x)>0$ for all $x\in(0,x_2)$ and $f'(x)<0$ for all $x\in(x_2,x_1)$, which  gives 
$$\min_{[0,5]}f=\min\left\{f(0),f\left(x_{1}\right)\right\}=\min\{0,0.0226...\}=0$$
and we are done!
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