Prove inequality $\arccos \left( \frac{\sin 1-\sin x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$ 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:

 A: 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!
A: Assuming radians: This is a good question. Using $\sin a - \sin b = 2\sin \frac{a-b}2 \cos \frac{a+b}2$ gives:
$$ \frac 2{1-x} \sin \frac{1-x}2 \cos \frac{1+x}2 \le \cos \sqrt \frac {(1+x)^2-x}3 $$
Then using (1+x)/2=u:
$$ \frac 1u \sin (1-u) \cos u \le \cos \sqrt \frac {4u^2-2u+1}3$$
Then there is an identity you could use on the left side that I can't remember... 
You could also prove that:
$$ \sin 1 \le (1-x) \cos \sqrt \frac {1+x+x^2}3 +\sin x $$
By differentiating the right side, finding minima and showing that they are all >sin1, but this is equally as hard (if not harder) as expanding everything as series.
A: This is not a full answer, just a possible way to prove the inequality.
We use the following form of the inequality:
$$\text{sinc} \left(\frac{1-x}{2} \right) \cos \left(\frac{1+x}{2} \right)  \geq \cos \sqrt{\frac{1+x+x^2}{3}}$$
It makes sense to try infinite products, mainly because we get rid of the square root:
$$\text{sinc}(t)=\prod_{n=1}^\infty \left(1-\frac{t^2}{\pi^2 n^2} \right)$$
$$\cos (t)=\prod_{n=1}^\infty \left(1-\frac{t^2}{\pi^2 (n-1/2)^2} \right)$$
Thus, our inequality becomes:
$$\prod_{n=1}^\infty \left(1-\frac{(1-x)^2}{4\pi^2 n^2} \right) \left(1-\frac{(1+x)^2}{4\pi^2 (n-1/2)^2} \right) \geq \prod_{n=1}^\infty \left(1-\frac{1+x+x^2}{3\pi^2 (n-1/2)^2} \right)$$
Note that for $x<1$ every term in the infinite products is positive.
If, for example, every term of the product on the left is greater than every term of the product on the right, then our inequality is proven.
$$\left(1-\frac{(1-x)^2}{4\pi^2 n^2} \right) \left(1-\frac{(1+x)^2}{4\pi^2 (n-1/2)^2} \right) \geq^? 1-\frac{1+x+x^2}{3\pi^2 (n-1/2)^2}$$
After expanding and simplifying I obtained the following:
$$-\left(4 \pi^2 n(2n-3)+3(\pi^2-(1+x)^2) \right)(1-x)^2 \geq^? 0$$
And this is highly questionable, i.e. not correct for most cases. Remember, we are interested in the case $x < 1$, so:
$$4 \pi^2 n(2n-3)+3(\pi^2-(1+x)^2) \leq^? 0$$
$$4 \pi^2 n(2n-3)+3\pi^2 \leq^? 3 (1+x)^2$$
For $n=1$ we have a trivial inequality:
$$-\pi^2 < 3 (1+x)^2$$
But for $n \geq 2$ the inequality quickly stops working.
So this method probably doesn't prove anything.

On the other hand, we can compare the two term products on each side, i.e. prove that:
$$\prod_{n=k}^{k+1} \left(1-\frac{(1-x)^2}{4\pi^2 n^2} \right) \left(1-\frac{(1+x)^2}{4\pi^2 (n-1/2)^2} \right) \geq \prod_{n=k}^{k+1} \left(1-\frac{1+x+x^2}{3\pi^2 (n-1/2)^2} \right)$$
If that fails, we can try $3$ product terms and so on. This is just some algebra that a CAS can take care of even for a large number of terms.

Update
I decided to rearrange the second product so the terms are of the same order in $x$ and $n$:
$$\prod_{n=1}^\infty \left(1-\frac{1+x+x^2}{3\pi^2 (n-1/2)^2} \right)=\prod_{k=1}^\infty \left(1-\frac{1+x+x^2}{3\pi^2 (2k-3/2)^2} \right) \left(1-\frac{1+x+x^2}{3\pi^2 (2k-1/2)^2} \right)$$
Still, we have the same relation: only the first terms of the products obey the inequality, while all the rest seem to break it.
Below you can see the plot for:
$$ \dfrac{\left(1-\frac{(1-x)^2}{4\pi^2 n^2} \right) \left(1-\frac{(1+x)^2}{4\pi^2 (n-1/2)^2} \right) }{ \left(1-\frac{1+x+x^2}{3\pi^2 (2n-3/2)^2} \right) \left(1-\frac{1+x+x^2}{3\pi^2 (2n-1/2)^2} \right) }$$
For $x \in [0,1]$ and $n=1,2,3, \dots$.


I see no way to prove that the product of all terms for $n \geq 2$ is still closer to $1$ than the first term (despite numerical evidence), so this way doesn't seem to work as a proof of the original inequality.

A: To prove the inequality, it is enough to show that the inequality becomes equality at a single point (namely $x=1$).
To see this, proceed by contradiction. Suppose that the inequality does not hold at some point $x\in[0,1]$. Then, since all functions considered are continuous, it follows it doesn't hold on an interval. On the other hand, by a straightforward calculation, there exist points $y<1<z$ such that the inequality holds at $x=y$ and $x=z$ with strict inequality. Therefore, by the Intermediate Value Theorem there must be at least two points that make the inequality an equality.
In this direction, it might be useful to consider the inequality as an optimization problem, and prove that it can only have one optimizer (i.e. one that makes the inequality an equality). Both $\sin$ and $\cos$ are concave functions on $[0,1]$, and $\sqrt x$ is also concave on $[0,1]$. I imagine one can put these facts together into a solution...
