Proving $x+\sin x-2\ln{(1+x)}\geqslant0$ Question: Let $x&gt;-1$, 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.
$\def\e{\mathrm{e}}\def\peq{\mathrm{\phantom{=}}{}}$Caution: This proof uses following inequalities of explicit values:$$
2π - 2\ln(1 + 2π) &gt; 1, \quad \frac{2}{π^2} (π - \ln(1 + π)) &gt; \frac{1}{π},\\
\frac{8}{5} - 2\ln 5 &gt; \frac{4}{5} - \frac{3}{5} \left( π + \arccos \frac{3}{5} \right).
$$
(For the sake of completeness, rigorous proofs of these three inequalities are given at the end.)

Case 1: $-1 &lt; x &lt; 0$. This case is easy to prove because $x &gt; \ln(1 + x)$ and $x &lt; \sin x$.

Case 2: $x \geqslant 2π$. Since $x + \sin x - 2\ln(1 + x) \geqslant x - 2\ln(1 + x) - 1$ and $x - 2\ln(1 + x) - 1$ is increasing for $x \geqslant 1$, then$$
x + \sin x - 2\ln(1 + x) \geqslant x - 2\ln(1 + x) - 1 \geqslant 2π - 2\ln(1 + 2π) - 1 &gt; 0.
$$

Case 3: $0 \leqslant x \leqslant π$. First, define $f_1(x) = \dfrac{2}{x^2} (x - \ln(1 + x))$, then$$
f_1'(x) = -\dfrac{2}{x^3} \left( \dfrac{(x + 2)x}{x + 1} - \ln(1 + x) \right).
$$
Define $f_2(x) = \dfrac{(x + 2)x}{x + 1} - \ln(1 + x)$, then $f_2'(x) = \dfrac{x^2 + x + 1}{(x + 1)^2} &gt; 0$. Thus $f_2(x) \geqslant f_2(0) = 0$, which implies $f_1'(x) \leqslant 0$ and $f_1(x) \geqslant f_1(π)$, i.e.$$
x - 2\ln(1 + x) \geqslant \frac{2}{π^2} (π - \ln(1 + π)) x^2 - x. \quad \forall x \in [0, π]
$$

Next, define $f_3(x) = \dfrac{x^2}{π} - x + \sin x$. To prove that $f_3(x) \leqslant 0$ for $0 &lt; x \leqslant π$, note that $f_3(π - x) = f_3(x)$, thus it suffices to prove for $0 &lt; x \leqslant \dfrac{π}{2}$. Because $\cos x$ is concave on $\left[ 0, \dfrac{π}{2} \right]$, then$$
\cos x \geqslant \frac{\cos\dfrac{π}{2} - \cos 0}{\dfrac{π}{2} - 0}(x - 0) + \cos 0 = -\frac{2}{π}x + 1, \quad \forall x \in \left[0, \frac{π}{2} \right]
$$
which implies $f_3'(x) = \dfrac{2}{π}x - 1 + \cos x \geqslant 0$ and $f_3(x) \geqslant f_3(0) = 0$. Thus$$
-\sin x \leqslant \frac{x^2}{π} - x. \quad \forall x \in [0, π]
$$

Therefore for $0 \leqslant x \leqslant π$,$$
x - 2\ln(1 + x) \geqslant \frac{2}{π^2} (π - \ln(1 + π)) x^2 - x \geqslant \frac{x^2}{π} - x \geqslant -\sin x,
$$
i.e. $x + \sin x - 2\ln(1 + x) \geqslant 0$.

Case 4: $π &lt; x &lt; 2π$. Note that $f_4(x) = x - 2\ln(1 + x)$ is convex on $(π, 2π)$ and $f_5(x) = -\sin x$ is concave on $(π, 2π)$, thus for $π &lt; x &lt; 2π$,$$
f_4(x) \geqslant f_4'(4)(x - 4) + f_4(4) = \frac{3}{5} x + \left( \frac{8}{5} - 2 \ln 5 \right),
$$\begin{align*}
f_5(x) &amp;\leqslant f_5'\left( π + \arccos\frac{3}{5} \right) \left( x - \left( π + \arccos\frac{3}{5} \right) \right) + f_5\left( π + \arccos\frac{3}{5} \right)\\
&amp;= \frac{3}{5}x + \left( \frac{4}{5} - \frac{3}{5} \left( π + \arccos\frac{3}{5} \right) \right),
\end{align*}
which implies $f_4(x) \geqslant f_5(x)$, i.e. $x + \sin x - 2\ln(1 + x) \geqslant 0$.

To prove the three inequalities at the beginning, the following three identities are needed:$$
π = 2\sum_{n = 0}^∞ \frac{n!}{(2n + 1)!!}, \quad \e = \sum_{n = 0}^∞ \frac{1}{n!}, \quad \ln\frac{1 + x}{1 - x} = 2\sum_{n = 0}^∞ \frac{x^{2n + 1}}{2n + 1},\\
\arccos(1 - x) = \sqrt{2x} \sum_{n = 0}^∞ \frac{(2n - 1)!!}{(2n)!!}·\frac{x^n}{(2n + 1) 2^n},
$$
where $|x| &lt; 1$. The first one see Wiki. The second one is well-known. The third one can be proved by noting that$$
\ln\frac{1 + x}{1 - x} = \ln(1 + x) - \ln(1 - x),\\
(\ln(1 + x))' = \frac{1}{1 + x} = \sum_{n = 0}^∞ (-x)^n, \quad (\ln(1 - x))' = -\frac{1}{1 - x} = -\sum_{n = 0}^∞ x^n.
$$
The fourth one can be proved by noting that$$
(\arccos(1 - x))' = -\frac{1}{\sqrt{2x}}·\frac{1}{\sqrt{1 - \dfrac{x}{2}}} = -\frac{1}{\sqrt{2x}} \sum_{n = 0}^∞ \binom{-\frac{1}{2}}{n} \left( -\frac{x}{2} \right)^n.
$$

Now, to prove that $2π - 2\ln(1 + 2π) &gt; 1$, note that $2π - 1 &gt; 5$ and$$
2\ln(1 + 2π) &lt; 5 \Longleftrightarrow (1 + 2π)^2 &lt; \e^5.
$$
Since $(1 + 2π)^2 &lt; 9^2 &lt; 2.5^5 &lt; \e^5$, then $2π - 2\ln(1 + 2π) &gt; 1$.

Next, note that$$
\frac{2}{π^2} (π - \ln(1 + π)) &gt; \frac{1}{π} \Longleftrightarrow \frac{\ln(1 + π)}{π} &lt; \frac{1}{2}.
$$
Define $f_6(x) = \dfrac{\ln(1 + x)}{x}$, then $f_6'(x) = \dfrac{1}{x^2} \left( \dfrac{x}{1 + x} - \ln(1 + x) \right).$ Define $f_7(x) = \dfrac{x}{1 + x} - \ln(1 + x)$, then $f_7'(x) = -\dfrac{x}{(1 + x)^2} \leqslant 0$, which implies $f_7(x) \leqslant f_7(0) = 0$ for $x \geqslant 0$. Thus $f_6'(x) \leqslant 0$ for $x \geqslant 0$, which implies$$
\frac{\ln(1 + π)}{π} = f_6(π) \leqslant f_6(3) = \frac{\ln 4}{3}.
$$
Since $\dfrac{\ln 4}{3} &lt; \dfrac{1}{2} \Leftrightarrow \e^3 &gt; 16$ and $\e^3 &gt; \left( \dfrac{8}{3} \right)^3 &gt; 16$, then $\dfrac{2}{π^2} (π - \ln(1 + π)) &gt; \dfrac{1}{π}$.

Finally, note that$$
\frac{8}{5} - 2\ln 5 &gt; \frac{4}{5} - \frac{3}{5} \left( π + \arccos \frac{3}{5} \right) \Longleftrightarrow 3\left( π + \arccos\frac{3}{5} \right) + 4 &gt; 10\ln 5.
$$
Since$$
π = 2\sum_{n = 0}^∞ \frac{n!}{(2n + 1)!!} &gt; 2\sum_{n = 0}^6 \frac{n!}{(2n + 1)!!} = \frac{141088}{45045},
$$\begin{align*}
\arccos\frac{3}{5} &amp;= \arccos\left( 1 - \frac{2}{5} \right) = \frac{2}{\sqrt{5}} \sum_{n = 0}^∞ \frac{(2n - 1)!!}{(2n)!!}·\frac{\left( \dfrac{2}{5} \right)^n}{(2n + 1) 2^n}\\
&amp;&gt; \frac{2}{\sqrt{5}} \sum_{n = 0}^1 \frac{(2n - 1)!!}{(2n)!!}·\frac{1}{(2n + 1) 2^n} \left( \frac{2}{5} \right)^n = \frac{31}{15\sqrt{5}},
\end{align*}\begin{align*}
\ln 5 &amp;= \ln\frac{1 + \dfrac{2}{3}}{1 - \dfrac{2}{3}} = 2\sum_{n = 0}^∞ \frac{1}{2n + 1} \left( \frac{2}{3} \right)^{2n + 1}\\
&amp;&lt; 2\sum_{n = 0}^5 \frac{1}{2n + 1} \left( \frac{2}{3} \right)^{2n + 1} + 2\sum_{n = 6}^∞ \frac{1}{13} \left( \frac{2}{3} \right)^{2n + 1}\\
&amp;= 2\sum_{n = 0}^5 \frac{1}{2n + 1} \left( \frac{2}{3} \right)^{2n + 1} + \frac{2}{13}·\frac{1}{1 - \left( \dfrac{2}{3} \right)^2}·\left( \frac{2}{3} \right)^{13} = \frac{6427830866}{7979586615},
\end{align*}
then\begin{align*}
&amp;\peq 3\left( π + \arccos\frac{3}{5} \right) + 4 &gt; 3\left( \frac{141088}{45045} + \frac{31}{15\sqrt{5}} \right) + 4\\
&amp;&gt; 10·\frac{6427830866}{7979586615} &gt; 10\ln 5.
\end{align*}
let's make several segments to prove it:


$x\ge \dfrac{3\pi}{2},x+\sin{x}-2\ln{(1+x)}\ge x-1-2\ln{(1+x)}=h(x)\ge 0 (h'&gt;0)$ 
$ \pi &lt;x &lt; \dfrac{3\pi}{2},x-2\ln{(1+x)}&gt;\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&gt;0 $
$0\le x\le\pi$, we only find $f_{max}$, so the min is $f(0)$ and $f(\pi),f(0)=0,f(\pi)&gt;0$,
$-1&lt;x&lt;0$,it is easy.