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.

Solution 1:

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.

Solution 2:

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

Case 1: $-1 < x < 0$. This case is easy to prove because $x > \ln(1 + x)$ and $x < \sin x$.

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$$ x + \sin x - 2\ln(1 + x) \geqslant x - 2\ln(1 + x) - 1 \geqslant 2π - 2\ln(1 + 2π) - 1 > 0. $$

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 $f_2'(x) = \dfrac{x^2 + x + 1}{(x + 1)^2} > 0$. 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 $0 < x \leqslant π$, note that $f_3(π - x) = f_3(x)$, thus it suffices to prove for $0 < x \leqslant \dfrac{π}{2}$. 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: $π < x < 2π$. 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 $π < x < 2π$,$$ f_4(x) \geqslant f_4'(4)(x - 4) + f_4(4) = \frac{3}{5} x + \left( \frac{8}{5} - 2 \ln 5 \right), $$\begin{align*} f_5(x) &\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)\\ &= \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. $x + \sin x - 2\ln(1 + x) \geqslant 0$.

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

Next, note that$$ \frac{2}{π^2} (π - \ln(1 + π)) > \frac{1}{π} \Longleftrightarrow \frac{\ln(1 + π)}{π} < \frac{1}{2}. $$ 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 $\dfrac{\ln 4}{3} < \dfrac{1}{2} \Leftrightarrow \e^3 > 16$ and $\e^3 > \left( \dfrac{8}{3} \right)^3 > 16$, then $\dfrac{2}{π^2} (π - \ln(1 + π)) > \dfrac{1}{π}$.

Finally, note that$$ \frac{8}{5} - 2\ln 5 > \frac{4}{5} - \frac{3}{5} \left( π + \arccos \frac{3}{5} \right) \Longleftrightarrow 3\left( π + \arccos\frac{3}{5} \right) + 4 > 10\ln 5. $$ Since$$ π = 2\sum_{n = 0}^∞ \frac{n!}{(2n + 1)!!} > 2\sum_{n = 0}^6 \frac{n!}{(2n + 1)!!} = \frac{141088}{45045}, $$\begin{align*} \arccos\frac{3}{5} &= \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}\\ &> \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 &= \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}\\ &< 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}\\ &= 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*} &\peq 3\left( π + \arccos\frac{3}{5} \right) + 4 > 3\left( \frac{141088}{45045} + \frac{31}{15\sqrt{5}} \right) + 4\\ &> 10·\frac{6427830866}{7979586615} > 10\ln 5. \end{align*}