Prove inequality $\arccos \left( \frac{\sin 1-\sin x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$
We have to show
$$\text{sinc} \left(\frac{1-x}{2} \right) \cos \left(\frac{1+x}{2} \right) \geq (1-\frac{(1-x)^2}{24}) \cos \left(\frac{1+x}{2} \right) \geq \cos \sqrt{\frac{1+x+x^2}{3}}$$
With the classic series for $\sin x$ and $0\leq x\leq 1$ it's clear that $$\text{sinc} \left(\frac{1-x}{2} \right) \geq 1-\frac{(1-x)^2}{24} $$
and therefore with a simple change of the first inequality it's now left
$$\frac{(1-x)^2}{24} \cos \left(\frac{1+x}{2} \right) \leq \cos \left(\frac{1+x}{2} \right)-\cos \sqrt{\frac{1+x+x^2}{3}}$$
or with $a:=\frac{1+x}{2}$ and $b:=\sqrt{\frac{1+x+x^2}{3}}$ and therefore $b\geq a$ it's
$$\frac{b^2-a^2}{2}\cos a \leq \cos a - \cos b$$
An equivalent inequality for this is $\int\limits_a^b (\sin x -x\cos a)dx \geq 0$.
$(A)\enspace$ Numerical proof with $0.5\leq a<b\leq 1$ for $\int\limits_a^b (\sin x -x\cos a)dx \geq 0$:
Increasing of $\sin x -x\cos a$: $(\sin x -x\cos a)'=\cos x - \cos a>0$ for $0\leq x<a$.
Decreasing with $a<x\leq b$.
Be $c:=\arccos(\sin(1))=0.570796…$ which means $\sin 1-1\cdot\cos c=0$.
(1) $\enspace c<a\leq 1$: $\enspace \sin x-x \cos a>0 \enspace$ for $\enspace a<x\leq b$
(2) $\enspace \frac{1}{2}\leq a\leq c$:
$\hspace{8mm}$ For every $a$ exists exactly one solution for $\sin x-x\cos a=0 \enspace$ when $\enspace a\leq x\leq b$ .
Definition: Be $x_0$ with $\sin x_0-x_0\cos \frac{1}{2}=0$
Because of $\enspace b=\sqrt{\frac{1-2a+4a^2}{3}}\enspace$ it’s $$\max\{b|\frac{1}{2}\leq a\leq c\}=\sqrt{\frac{1-2c+4c^2}{3}}=0.62226498459…<\frac{3}{4}<$$ $$<\min\{x\in[a;b]|\sin x-x \cos a=0 \text{ with }\frac{1}{2}\leq a\leq c\}=x_0=0.873…$$ => $\enspace \sin x-x \cos a>0$ for $a\leq x\leq b$
Therefore with (1)+(2) it’s $\int\limits_a^b (\sin x-x\cos a)dx \geq 0$ as expected.
In words: The integrand $\sin x-x\cos a$ is always positiv within the valid value area and therefore the integral too.
$(B)\enspace$ A non-numerical proof using the first part of the explanations above:
We have to show that $\int\limits_a^b (\sin x-x\cos a)dx \geq 0$ .
This is true if $\enspace\sin x-x\cos a\geq 0\enspace$ for $\enspace a\leq x\leq b$ .
$\sin x-x\cos a\enspace$ is decreasing for $\enspace a<x<b\enspace$ because of $\enspace\displaystyle \frac{d}{dx}(\sin x-x\cos a)<0\enspace$ and therefore is
$\hspace{1cm}$ $\min\{\sin x-x\cos a|a\le x\le b\}=\sin b-b\cos a$
$\hspace{1cm}$ for $\enspace 0.5\leq a\leq b\leq 1\enspace$ with $\enspace b=\sqrt{\frac{1-2a+4a^2}{3}}$ .
$=>\enspace$ It has to be shown that $\enspace\displaystyle \cos a<\frac{\sin b}{b}\enspace$ e.g. by proving $\enspace\displaystyle\cos a<1.1-0.4 a<\frac{\sin b}{b}$ .
The left side is clear for $\enspace\displaystyle\frac{1}{2}\leq a\leq 1\enspace$ and the right side can be better handled if $\enspace a\enspace$ is substituted by $\enspace\displaystyle\frac{1}{4}(1+\sqrt{3}\sqrt{(2b)^2-1})\enspace$ with $\enspace\displaystyle\frac{1}{\sqrt{3}}\leq b\leq 1\enspace$ so that we can simplify e.g. $\enspace\displaystyle 1-0.1\sqrt{3}\sqrt{(2b)^2-1}<1.15-0.4 b<\frac{\sin b}{b}$ .
It's $\enspace\displaystyle 1-0.1\sqrt{3}\sqrt{(2x)^2-1}<1.15-0.4 x\enspace$ true for $\enspace\displaystyle |x-1.5|<\frac{1}{4}\sqrt{15}\enspace$ which includes $\enspace\displaystyle \frac{1}{\sqrt{3}}\leq x\leq 1$ .
To verify $\enspace\displaystyle \cos a<\frac{\sin b}{b}\enspace$ we can use the classical series of $\enspace\cos\enspace$ and $\enspace\sin\enspace$ and get the following inequalities which have to be proved:
$\hspace{8mm}\displaystyle\cos x<1-\frac{x^2}{2}+\frac{x^4}{24}<1.1-0.4 x\enspace$ for $\enspace\displaystyle\frac{1}{2}\leq x\leq 1\enspace$ and
$\hspace{8mm}\displaystyle 1.15-0.4 x<1-\frac{x^2}{6}<\frac{\sin x}{x}\enspace$ for $\enspace\displaystyle\frac{1}{\sqrt{3}}\leq x\leq 1\enspace$
$(1)\enspace\displaystyle \cos x<1-\frac{x^2}{2}+\frac{x^4}{24}\enspace$ for $\enspace\displaystyle\frac{1}{2}\leq x\leq 1$ :
$\hspace{8mm}$ This is true with $\enspace\displaystyle\cos x=\sum\limits_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}$
$\hspace{8mm}$ because of $\enspace\displaystyle\frac{x^{2k}}{(2k)!}-\frac{x^{2k+2}}{(2k+2)!}>0\enspace$ for $\enspace k\in\mathbb{N}$ .
$(2)\enspace\displaystyle 1-\frac{x^2}{2}+\frac{x^4}{24}<1.1-0.4 x\enspace$ for $\enspace\displaystyle\frac{1}{2}\leq x\leq 1$ :
$\hspace{8mm}$ This is true for $\enspace\displaystyle x:=\min{x}=\frac{1}{2}$
$\hspace{8mm}$ and because of $\enspace\displaystyle\frac{d}{dx}(1-\frac{x^2}{2}+\frac{x^4}{24})<\frac{d}{dx}(1.1-0.4 x)<0$ .
$(3)\enspace\displaystyle 1.15-0.4 x<1-\frac{x^2}{6}\enspace$ for $\enspace\displaystyle\frac{1}{\sqrt{3}}\leq x\leq 1\enspace$ :
$\hspace{8mm}$ This is true for $\enspace\displaystyle x:=\min{x}=\frac{1}{\sqrt{3}}$
$\hspace{8mm}$ and because of $\enspace\displaystyle\frac{d}{dx}(1.15-0.4 x)<\frac{d}{dx}(1-\frac{x^2}{6})<0$ .
$(4)\enspace\displaystyle 1-\frac{x^2}{6}<\frac{\sin x}{x}\enspace$ for $\enspace\displaystyle\frac{1}{\sqrt{3}}\leq x\leq 1\enspace$ :
$\hspace{8mm}$ This is true with $\enspace\displaystyle \frac{\sin x}{x} =\sum\limits_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k+1)!}$
$\hspace{8mm}$ because of $\enspace\displaystyle\frac{x^{2k}}{(2k+1)!}-\frac{x^{2k+2}}{(2k+3)!}>0\enspace$ for $\enspace k\in\mathbb{N}$ .
With the verification of $(1)$ to $(4)$ the proof is completed.
A summary of the steps of $(B)$ .
$\displaystyle sinc(\frac{1-x}{2})\cos(\frac{1+x}{2})\geq \cos\sqrt{\frac{1+x+x^2}{3}}$ is verified by proofs for $(1)$ and $(2)$ .
$(1)\enspace$ $\displaystyle sinc(\frac{1-x}{2})\geq 1-\frac{(1-x)^2}{24}\enspace$ (proof with series expansion)
$(2)\enspace$ $\displaystyle (1-\frac{(1-x)^2}{24})\cos(\frac{1+x}{2})\geq\cos\sqrt{\frac{1+x+x^2}{3}}\enspace$ (verified by the proof for $(3)$)
With $\enspace\displaystyle a:=\frac{1+x}{2}\in [\frac{1}{2};1]\enspace$ and $\enspace\displaystyle b:=\sqrt{\frac{1+x+x^2}{3}}\in [\frac{1}{\sqrt{3}};1]\enspace$ point $(2)$ changes to
$(3)\enspace$ $\displaystyle\int\limits_a^b (\sin x-x\cos a)dx\geq 0$ .
Because of $\enspace\min(\sin x-x\cos a)|_{a\leq x\leq b}=\sin b-b\cos a\enspace$ (proof by derivation) point $(3)$
is verified by the proof for $\enspace\displaystyle\cos a<\frac{\sin b}{b}\enspace$, points $(4)$ till $(8)$ .
$\displaystyle\frac{1}{2}\leq x\leq 1$ :
$(4)\enspace$ $\displaystyle \cos x<1-\frac{x^2}{2}+\frac{x^4}{24}\enspace$ (proof with series expansion)
$(5)\enspace$ $\displaystyle 1-\frac{x^2}{2}+\frac{x^4}{24}<1.1-0.4x\enspace$ (proof with derivation)
$\displaystyle\frac{1}{\sqrt{3}}\leq x\leq 1$ :
$(6)\enspace$ $\displaystyle 1-0.1\sqrt{3}\sqrt{(2x)^2-1}<1.15-0.4 x\enspace$ (proof with solving the quadratic equation)
$(7)\enspace$ $\displaystyle 1.15-0.4 x<1-\frac{x^2}{6}\enspace$ (proof with derivation)
$(8)\enspace$ $\displaystyle 1-\frac{x^2}{6}<\frac{\sin x}{x}\enspace$ (proof with series expansion)
Note: $(6)$ and $(7)$ can be put together; I haven't, for a better overfiew.
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!
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.