TITLE: Prove inequality $\arccos \left( \frac{\sin 1-\sin x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$
QUESTION [43 upvotes]: I was trying to figure out if the following function can serve as a mean (see mean value theorem):
$$\arccos \left( \frac{\sin y-\sin x}{y-x} \right)$$
And turns out that for $x,y \leq \pi$ it does serve as a mean admirably.
But then I've noticed that for $0<x<1$ the following two functions are very close (see the picture):


Now how would you prove:
$$\arccos \left( \frac{\sin 1-\sin x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$$

It's probably easier to consider another equivalent inequality:
$$\frac{\sin 1-\sin x}{1-x}  \geq \cos \sqrt{\frac{1+x+x^2}{3}}$$
Or even:
$$ \text{sinc} \left(\frac{1-x}{2} \right) \cos \left(\frac{1+x}{2} \right)  \geq \cos \sqrt{\frac{1+x+x^2}{3}}$$
We could use Taylor series, but that's too cumbersome in my opinion.
Another way would be Mean value theorem itself, but I encounter the same problem.

Is there a simple way to prove this inequality?

My calculus is not as sharp as it used to be (just kidding, it was never sharp).

Edit
Just to confirm (numerically) that the inequality holds, here is the plot of the difference between the two functions:

REPLY [6 votes]: Geometric Explanation “Why $3$ ?”
(Not an Answer)  

In order to explain why the inequality could be true, we have:
$$ x^3-1=(x-1)(1+x+x^2) \Rightarrow \cos\sqrt{\frac{1+x+x^2}{3}}=\cos\sqrt{\frac{1}{\color{red}{3}}\frac{x^{\color{red}{3}}-1}{x-1}} $$
And by considering the general case $\space x^n-1=(x-1)(1+x+x^2+\cdots+x^{n-1})$, Let:
$$
\begin{align}
& \color{red}{f_{\alpha}(x)} = \frac{\sin(1)-\sin(x)}{1-x}-\cos\sqrt{\frac{1}{\color{red}{\alpha}}\frac{x^{\color{red}{\alpha}}-1}{x-1}} \quad\colon\space \alpha \ge 1, \space \alpha \in \mathbb{R} \\[2mm]
& \qquad \Rightarrow \space f_{\alpha}(0) = \lim_{x\rightarrow0}f_{\alpha}(x) = \sin(1)-\cos\left(1/\sqrt{\alpha}\right) \\
& \qquad \rightarrow \space \text{for}\space f_{\alpha}(0)=0 \Rightarrow \alpha=1/\arccos^2\left(\sin(1)\right) \\[2mm]
& \qquad \space\&\space\space \space f_{\alpha}(1)= \lim_{x\rightarrow1}f_{\alpha}(x) = 0 \quad \left\{\text{for all}\space\alpha\right\} \\
& \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}}
\end{align}
$$
Also:
$$
\begin{align}
& \color{red}{f'_{\alpha}(x)} = \small \frac{\sin(1)-\sin(x)-(1-x)\cos(x)}{(1-x)^2}-\frac{1-x^\alpha-\alpha(1-x)x^{\alpha-1}}{2\alpha(1-x)^2}\sqrt{{\alpha}\frac{x-1}{x^\alpha-1}}\,\sin\sqrt{\frac{1}{\alpha}\frac{x^\alpha-1}{x-1}} \\[2mm]
& \qquad \Rightarrow \space f'_{\alpha}(0) = \lim_{x\rightarrow0}f'_{\alpha}(x) = \sin(1)-1+\frac{1}{\sqrt{\alpha}}\,\sin\left(1/\sqrt{\alpha}\right) \\[2mm]
& \qquad \space\&\space\space \space f'_{\alpha}(1)= \lim_{x\rightarrow1}f'_{\alpha}(x) = \color{red}{\frac{\sin(1)}{4}\,(\alpha-3)} \\
& \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}}
\end{align}
$$



By swiping $\alpha$ from $1$ towards $\infty$, considering the range $x\in[0,1]$:   
${\bf1}$. Starting by $\space f_1(x)=\left[\sin(1)-\sin(x)\right]/[1-x]-\cos(1)$,   a well defined function,  
$\qquad$ smooth decreasing curve from $(0,\sin1-\cos1)$ to $(1,0)$, no $x$ axis intersection (no zeros).   
${\bf2}$. Smooth adjustment in the decreasing behavior while $\alpha$ increases.   
${\bf3}$. At a certain value $1\lt\alpha_0\lt1/\arccos^2\left(\sin(1)\right)$, $f_{\alpha}(x)$ will stop being completely decreasing,  
$\qquad$ start to have an increasing part, causing intersection(s) with $x$ axis (at least one zero).   
${\bf4}$. Between $\alpha_0\lt\alpha\lt1/\arccos^2\left(\sin(1)\right)$, $f_{\alpha}(x)$ surly has at least one zero.   
${\bf5}$. For $\alpha\gt1/\arccos^2\left(\sin(1)\right)\Rightarrow f_{\alpha}(0)\lt0$,  
$\qquad$ and the function should stop intersecting with $x$ axis.   
${\bf6}$. When $\alpha\rightarrow\infty\space\Rightarrow(x^\alpha-1)/(x-1)\approx1/(1-x) \Rightarrow f_{\alpha}(x)\approx1\frac{\sin(1)-\sin(x)}{1-x}-\cos\left(\frac{1/\alpha}{1-x}\right)$.   

And from the definition of $f'_{\alpha}(x)$, we can calculate the value of $\alpha_0$ that stops the completely decreasing behavior and create a horizontal tangent at the end of the interval $x\in[0,1]$:
$$ f'_{\alpha_0}(1)=0 \space\Rightarrow\space \frac{\sin(1)}{4}\left(\alpha_0-3\right)=0 \space\Rightarrow\space \color{red}{\alpha_0=3} $$
The inequality is really interesting because it shows the maximum $\alpha$ of $\left\{f_{\alpha}(x)\ge0\colon x\in[0,1]\right\}$  
As-well-as, for $\left\{3\lt\alpha\lt1/\arccos^2\left(\sin(1)\right)\right\}$ we have $\left\{f_{\alpha}(0)>0 \space\&\space f'_{\alpha}(1)>0\right\}$, Thus:  
The function $f_{\alpha}(x)$ have an odd number of zeros in the range $x\in[0,1]$.  
In fact, if we can argue to replace the statement "at least one zero" with "at most one zero" for $3\lt\alpha\lt1/\arccos^2\left(\sin(1)\right)$, then the equality is proved!