# 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 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:

• If I'm reading it correctly, the graph you have seems to suggest that $\arccos\left(\frac{\sin(1) - \sin x)}{1-x}\right) \ge \sqrt{\frac{1+x+x^2}{4}}$. Is this the inequality you want to show? Or is the graph mislabeled? (Or am I missing something?)
– user88319
Aug 25, 2016 at 15:50
• @Strants, I mislabeled it, yes. Thank you! Aug 25, 2016 at 15:56
• @xpaul, I plotted every mean I know for $x$ and $1$ and then plotted even more means. Aug 25, 2016 at 16:07
• It is interesting. Aug 25, 2016 at 16:10
• @xpaul, I just moved on from using boring old power means to using Cauchy mean value theorem to generate functional means. This is the start. I'm sure I'll get more interesting inequalities soon. Aug 25, 2016 at 16:13

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.

• It's not clear to me how you prove $$\frac{b^2-a^2}{2}\cos a \leq \cos a - \cos b$$ I'll get it in time, but it would be nice if you elaborated Aug 26, 2016 at 21:50
• @Yuriy S: I have changed the argumentation. The numerical way doesn't look nice but it works (if I haven't again a mistake). Aug 27, 2016 at 20:03
• @Yuriy S: The decisive point of my proof is, that $\sin x-x\cos a>0$ for $\frac{1}{2}\leq a\leq x\leq b\leq 1$ taking into account that $b$ is given by $a$. Is there a mistake ? Jan 2, 2017 at 18:08
• user90369, it's good, but you use numerical argument Jan 2, 2017 at 19:05
• @Yuriy S: Yes. :-) Hint: $\sin x-x\cos a$ is not arbitrarily close to zero, it's $\sin x-x\cos a\geq \sin \frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}\cos \frac{1}{2}>0.039$ for the whole value range. (I haven't thought about this in August but) I think that's the easiest way to proof the positivity of the integral. When I have time I will do this but maybe you would have calculated this faster than me. Jan 2, 2017 at 23:44

## Geometric Explanation “Why $3$ ?”

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.

• The function on the RHS in the last expression has a very nice series around $x=1$: $$\sin 1-\frac{1}{24} (\sin 1-\cos 1)(1-x)^3+\frac{1}{48} \cos 1 (1-x)^4+O((1-x)^5)$$ See wolframalpha.com/input/… Aug 25, 2016 at 19:00
• I like the first sentence: Assuming radians: This is a good question Dec 30, 2016 at 1:38

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.

• Clever idea... but the inequality shouldn't fail with infinite products... we just need the right manipulation of the capital pi notation. Perhaps induction on the final part of your answer? Aug 27, 2016 at 18:22
• @BenLaurense, I'm not sure. I've succesfully proved another inequality using infinite products and thought it will be the same for this one. But everything I've tried (i.e. several terms together) works only for $n=1$ and fails for $n \geq 2$. Aug 27, 2016 at 18:24
• My intuition says: we know it's true, so it must be possible to prove it through products (right?). But if you've tried all the obvious manipulations then it's probably something weird like multiplying together every fifth term in the left product and every third term in the right product. Aug 27, 2016 at 18:35
• @BenLaurense, I checked again and it seems like the inequality really only works for $n=1$, but the ratio between the first terms on the l.h.s and r.h.s is large enough that even if we multiply by all the ratios for $n>1$ it still stays greater than $1$, thus the inequality is true but can't be proven with the infinite product, since only 1 term obeys the same inequality Sep 11, 2017 at 14:30

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...

• Is there a misunderstanding? The second paragraph states why the first line is correct. If you see a mistake there, please tell me. Jan 1, 2017 at 2:14
• Sorry, I misinterpret the sentence due to my poor English :( It's correct. Jan 1, 2017 at 3:26