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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.

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    $\begingroup$ For $x\leqslant0$ we can use $\ln(1+x)\leqslant x$, and for $x\geqslant2e-1$, $\ln(x+1)\leqslant\frac{x-1}2$. Seems like $2e-1$ was a good guess, it is pretty close to the actual zero of $\ln(x+1)-\frac{x-1}2$. $\endgroup$ Jun 22, 2014 at 7:59
  • $\begingroup$ Taking the derivative is a dead end, since the function is not monotonous on $(-1,\infty)$. $\endgroup$
    – Lucian
    Jun 22, 2014 at 8:05
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    $\begingroup$ This is really tricky with the almost-zero around $4$. $\endgroup$ Jun 22, 2014 at 8:07
  • $\begingroup$ Isn't $1-\frac{2}{1+x}=\frac{x-1}{1+x}$? $\endgroup$
    – Ellya
    Jun 22, 2014 at 8:28
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    $\begingroup$ This is the simplest looking question I have seen during my time here that nobody can answer in a simple and beautiful way for so long time. $\endgroup$
    – A.Γ.
    Oct 1, 2018 at 19:20

6 Answers 6

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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.

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  • $\begingroup$ +1, very natural and clean! Minor question: Do you use an approach similar to that @AlexFrancisco adopts to prove the third inequality in his first paragraph to prove $f(x_3)>0$? I suppose you used Taylor series expansion to show $f'(x_3)\in(0,k)$. If you use a different approach, I am interested to see it. Thank you. $\endgroup$
    – Hans
    Oct 3, 2018 at 8:25
  • $\begingroup$ @Hans : Thank you for your comment. To estimate $f'(x_3)$, I used Mathematica, in view of the fact that Mathematica can compute values of elementary and special functions to any degree of accuracy/precision. Another way to do it would be to use Taylor expansions and do all arithmetical calculations by hand; that I did not do. $\endgroup$ Oct 3, 2018 at 13:26
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    $\begingroup$ Could the person who down-voted this answer please explain why? It is baffling. $\endgroup$
    – Hans
    Oct 4, 2018 at 4:15
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$\def\e{\mathrm{e}}\def\peq{\mathrm{\phantom{=}}{}}$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. $x + \sin x - 2\ln(1 + x) \geqslant 0$.


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*}

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  • $\begingroup$ Could you please prove the third inequality in "Caution"? Thank you, Alex. $\endgroup$
    – Hans
    Oct 2, 2018 at 17:13
  • $\begingroup$ @Hans Actually there's no easy way to prove the third one apart from using well-established inequalities in numerical analysis. $\endgroup$ Oct 2, 2018 at 23:29
  • $\begingroup$ Are you referring to the interval arithmetics? For the sake of completeness, would you mind giving a sketch of the proof? It will be instructive for other readers, too. Thanks, Alex. $\endgroup$
    – Hans
    Oct 3, 2018 at 0:41
  • $\begingroup$ @Hans I added the proof for those inequalities, but these proofs are not at all interesting. $\endgroup$ Oct 3, 2018 at 4:09
  • $\begingroup$ +1! You may be right, but I would say they are very instructive. :-) I, for one, got the the first two inequalities but had a much more cumbersome proof for the third than yours. Excellent proof! Your approach for the main body is a bit tricky. $\endgroup$
    – Hans
    Oct 3, 2018 at 8:30
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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 $f(x) \ge 0$.

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

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 $f(x) \ge 0$.

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

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.

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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.

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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!

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    $\begingroup$ No, that is NO proof. You need to PROVE the inequality. We have no idea how accurate this $0.0226$ or $4.06268$ is until you prove it. Had that been counted as a proof, we could have just plot the graph with a decent computer and eye-ball the whole thing and be done with it, and dispense with all you have written above. $\endgroup$
    – Hans
    Sep 29, 2018 at 9:55
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    $\begingroup$ In case, you do not see what I typed in the chatroom, I copy my comment here. You have shown that there exists an $x_1\in [\pi, 5]$ such that $x_1$ is an interior minimum of $f$ in $[\pi,5]$. However, you need to show that indeed $f(x_1)>0$. By writing $f(x_1)=0.0226...$ you are claiming $f(x_1)\in [0.0226,0.0227)$. You also say $x_1=4.06268...$ equivalent to claiming $x_1\in[4.06268,4.06269)$. You have not provided the proofs for the two aforementioned intervals. Claiming these come from a calculator does not constitute a proof with explicit logical steps. $\endgroup$
    – Hans
    Sep 29, 2018 at 20:02
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    $\begingroup$ The key to the proof is $f(x_1)\ge0$, since if $f(x_1)<0$ the proposition fails. Yes, $f$ is a continuous and differentiable function, and all the other conclusions of your proof are true but it does not suffice for $f(x_1)>0$. The only place that property is ever mentioned is the last sentence claiming $f(x_1)\in[0.0226,0.0227)$. There is no derivation of that statement anywhere. Otherwise, can you point to the part that actually proves this claim? If you think it should be proved but do not want to write out the long proof, it would be better to explicitly say so in the answer. $\endgroup$
    – Hans
    Sep 30, 2018 at 9:48
  • 1
    $\begingroup$ @Hans I agree with Hans, you should do more careful calculations, a step cannot be simply replaced by numerics at some point. Otherwise, we should accept the WolframAlfa proof from the beginning. By the way, it is a continuous function there as well. $\endgroup$
    – A.Γ.
    Oct 1, 2018 at 19:12
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    $\begingroup$ I agree with Hans that it is not quite clear how $f(x_1)$ was computed or how it was shown that $f(x_1)>0$. Of course, this is a minor gap. $\endgroup$ Oct 2, 2018 at 21:26
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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.

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  • $\begingroup$ This is not right. Case 3 is not proved. If you want to claim the stationary point is unique and a local maximum, you need to show it. In Case 2, $g_{\text{min}}$ needs to be proved rigorously greater than $0$. $\endgroup$
    – Hans
    Sep 26, 2018 at 21:02

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