Proving $x+\sin x-2\ln{(1+x)}\geqslant0$

Question

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.

Answer 1

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.

Answer 2

let's make several segments to prove it:

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

2. $ \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 $

3. $0\le x\le\pi$, we only find $f_{max}$, so the min is $f(0)$ and $f(\pi),f(0)=0,f(\pi)>0$,

4. $-1<x<0$,it is easy.

Answer 3

Alternative proof:

Let $f(x) = x + \sin x - 2\ln (1+x)$.

We split into three cases:

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

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 $f(x) \ge 0$.

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

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 $f(x) \ge 0$.

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

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.

$\phantom{2}$

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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.