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

$\def\e{\mathrm{e}}\def\peq{\mathrm{\phantom{=}}{}}$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. $x + \sin x - 2\ln(1 + x) \geqslant 0$.

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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 $|x| < 1$. 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 $2π - 2\ln(1 + 2π) > 1$, 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*}

Answer 2

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.

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.

Answer 4

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!

Answer 5

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 6

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.