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.
 A: We need to show that 
\begin{equation*}
 f(x):=x+\sin x-2\ln{(1+x)}\ge0
\end{equation*}
for $x>-1$. We have 
\begin{equation*}
 f(x)\ge g(x):=x-1-2\ln{(1+x)},
\end{equation*}
$g$ is convex, $g(2\pi)>0$, and $g'(2\pi)>0$, so that $g>0$ on $[2\pi,\infty)$ and hence
\begin{equation*}
 f>0\quad\text{on}\quad[2\pi,\infty). \tag{1}
\end{equation*}
Next, $f''(x)=-\sin x+\frac2{(1+x)^2}$ is decreasing in $x\in[0,\pi/2]$ and hence $f''\ge f''(1/2)>0$ on $[0,1/2]$. Also, on $(-1,0)\cup[\pi,2\pi]$ we have $\sin\le0$ and hence $f''>0$. 
So, $f$ is convex on $(-1,1/2]$ and on $[\pi,2\pi]$. 
Next,
\begin{equation*}
 f''''(x)=\sin x+\frac{12}{(1+x)^4}>0\quad\text{for}\quad x\in[1/2,\pi], 
\end{equation*}
so that $f''$ is strictly convex on $[1/2,\pi]$ and hence has at most two roots in $[1/2,\pi]$. Since $f''(1/2)>0$, $f''(2)<0$, and $f''(\pi)>0$, we see that $f''$ changes sign exactly twice on $[1/2,\pi]$. So, for some $x_1$ and $x_2$ such that 
$$1/2<x_1<x_2<\pi,$$
\begin{equation*}
\text{$f$ is convex on $(-1,x_1]$ and on $[x_2,2\pi]$, and $f$ is concave on $[x_1,x_2]$. } 
\end{equation*}
Since $f(0)=0=f'(0)$, we have 
\begin{equation*}
 f\ge0\quad\text{on}\quad(-1,x_1]. \tag{2}
\end{equation*}
Let $x_3:=\frac{4063}{1000}\in[\pi,2\pi]\subset[x_2,2\pi]$ and $k:=278/10^6$. Then $f'(x_3)\in(0,k)$. Therefore and because $f$ is convex on $[x_2,2\pi]$, we have 
\begin{equation*}
 f(x)\ge f(x_3)+f'(x_3)(x-x_3)\ge f(x_3)+k(x-x_3)\ge f(x_3)+k(1/2-x_3)>0
\end{equation*}
for $x\in[x_2,x_3]$ and 
\begin{equation*}
 f(x)\ge f(x_3)+f'(x_3)(x-x_3)\ge f(x_3)>0
\end{equation*}
for $x\in[x_3,2\pi]$. So, 
\begin{equation*}
 f>0\quad\text{on}\quad[x_2,2\pi]. \tag{3}
\end{equation*}
Since $f$ is concave on $[x_1,x_2]$, it follows from (2) and (3) that 
\begin{equation*}
 f\ge0\quad\text{on}\quad[x_1,x_2]. \tag{4}
\end{equation*}
Finally, (1)--(4) yield 
\begin{equation*}
 f\ge0\quad\text{on}\quad(-1,\infty),
\end{equation*}
as desired. 
A: Alternative proof:
Let $f(x) = x + \sin x - 2\ln (1+x)$.
We split into three cases:

*

*$x\in (-1, 0]$:

We have $f'(x) = 1 + \cos x - \frac{2}{1+x} \le 1 + 1 - \frac{2}{1+x} = \frac{2x}{1+x} \le 0$.
As $f(0) = 0$, we have $f(x) \ge 0$.


*$x\in [0, \frac{3}{2}]$:

Since $\cos x \ge 1 - \frac{x^2}{2}$ for $x\in \mathbb{R}$,
we have $$f'(x) = 1 + \cos x - \frac{2}{1+x} \ge 1 + 1 - \frac{x^2}{2} - \frac{2}{1+x}
= \frac{x(4-x-x^2)}{2+2x} \ge 0$$
where we have used $4-x-x^2 \ge 4 - \frac{3}{2} - (\frac{3}{2})^2 > 0$.
As $f(0) = 0$, we have $f(x) \ge 0$.


*$x\in [\frac{3}{2}, \infty)$:

We have the following results. The proofs are given at the end.
Fact 1: It holds that $-\frac{3}{5}(x-4) + 2\ln 5 - 4 \ge 2\ln (1+x) - x$.
Fact 2: It holds that $\sin x \ge -\frac{3}{5}(x-4) + 2\ln 5 - 4$.
We are done.
$\phantom{2}$

Remarks: In the following proofs, we need to prove that $g(3/2) = \sin \tfrac{3}{2} + \tfrac{5}{2} - 2\ln 5 \ge 0$,
$g(\pi) = \frac{3}{5}\pi + \frac{8}{5} - 2\ln 5 \ge 0$ and
$g(\pi + \arccos \frac{3}{5}) = \tfrac{4}{5} + \tfrac{3}{5}\pi + \tfrac{3}{5}\arccos \tfrac{3}{5} - 2\ln 5 \ge 0$. One may use a calculator.
If one wants to prove it by hand, the proof is easy but annoying.
Proof of Fact 1: Denote $(\mathrm{LHS}-\mathrm{RHS})$ by $h(x)$. We have $h'(x) = \frac{2(x-4)}{5 + 5x}$.
Thus, $h(x)$ is non-increasing on $[\frac{3}{2}, 4]$, and non-decreasing on $[4, \infty)$.
As $h(4) = 0$, we have $h(x) \ge 0$. We are done.
Proof of Fact 2: Denote $(\mathrm{LHS}-\mathrm{RHS})$ by $g(x)$.
We have $g'(x) = \cos x + \frac{3}{5}$ and $g''(x) = -\sin x$. There are three possible cases:
i) $g(x)$ is concave on $[\frac{3}{2}, \pi]$.
Thus, we have $g(x) = g(\tfrac{\pi-x}{\pi - 3/2} \cdot \frac{3}{2} + \tfrac{x-3/2}{\pi - 3/2}\cdot \pi)
\ge \tfrac{\pi-x}{\pi - 3/2} g(3/2) + \tfrac{x-3/2}{\pi - 3/2}g(\pi) \ge 0$ on $[\frac{3}{2}, \pi]$
since $g(3/2)\ge 0$ and $g(\pi)\ge 0$.
ii) $g(x)$ is convex on $[\pi, 2\pi]$. Also, on $[\pi, 2\pi]$, $g'(x) = 0$ has a unique solution $x = \pi + \arccos\frac{3}{5}$.
Thus, $g(x) \ge g(\pi + \arccos \frac{3}{5}) \ge 0$ on $[\pi, 2\pi]$.
iii) If $x \ge 2\pi$, we have $g(x) \ge -1 + \frac{3}{5}(2\pi-4) - 2\ln 5 + 4
\ge -1 + \frac{3}{5}(2\pi-4) - 2\ln (\mathrm{e}^2) + 4 \ge 0$.
We are done.
A: let's make several segments to prove it:


*

*$x\ge \dfrac{3\pi}{2},x+\sin{x}-2\ln{(1+x)}\ge x-1-2\ln{(1+x)}=h(x)\ge 0 (h'>0)$ 

*$ \pi <x < \dfrac{3\pi}{2},x-2\ln{(1+x)}>\dfrac{3x+8}{5}-2ln5,g(x)=\dfrac{3x+8}{5}-2ln5+\sin{x},g'(x)=0$, we find a min value $g_{min}=\dfrac{3(\pi+\sin^{-1}{0.8})+8}{5}-2ln5=.0224>0 $

*$0\le x\le\pi$, we only find $f_{max}$, so the min is $f(0)$ and $f(\pi),f(0)=0,f(\pi)>0$,

*$-1<x<0$,it is easy.
A: If $g(x)=f(x+2\pi)-f(x),$ then 
$$g(x)=2\pi-2\ln \frac{x+2\pi+1}{x+1}$$
which is monotone increasing and positive at $x=0$. This means that, provided $f(x)\ge 0$ is shown for $x \in (-1,2\pi],$ then it follows that $f(x) \ge 0$ for any $x>-1.$
I don't have any way other than numerical estimates of the derivative of $f$ to obtain the min of $f$ on $(-1,2\pi],$ but it is $0$ at zero, since the only other possibility is at the second zero ($x \approx 4.06208$) of $f'(x)$ (a local min of $f$) and $f$ positive there (about $0.022628$). So if one is willing to live with this use of approximations to minimize $f$ on $(-1,2\pi]$ the inequality follows.
A: 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!
