inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ How can I prove the inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ ?
The derivative of $f(x):=\sqrt{\cos x}-\cos(\sin x)$ is very unpleasant, so the standard method is probably not be the right choice...
 A: Both sides being positive, we can raise each to the fourth power, to get
$$ \cos^2 x > \cos^4 \sin x $$
and then (as suggested by @Bacon), change variable to $u=\sin x$, which gives us
$$ 1-u^2 > \cos^4 u, \qquad u\in(0,1/\sqrt 2) $$
By Taylor's theorem $\cos u = 1 - \frac12 u^2 + \frac{\cos \xi}{4!}u^4$ for some $\xi\in(0,1/\sqrt2)$, which implies $\cos\xi>0$. Raising this to the fourth power gives us


*

*Terms only involving $1$ and $\frac12u^2$, namely $1 - 4\cdot\frac12u^2 +6\cdot\frac14u^4 - 4\cdot\frac18u^6+\frac1{16}u^8 $

*Further negative terms involving one or three factors of $-\frac12u^2$, which we can ignore

*Positive tems with exactly one $\frac{\cos\xi}{24}u^4$, namely $4\frac{\cos\xi}{24}u^4 + 12\frac{\cos\xi}{24\cdot 4}u^8$

*Fewer than $\binom42\cdot 3^2=54$ terms with two or more factors of $\frac{\cos\xi}{24}u^4$.


Thus we have
$$ \cos^4 u < 1 - 2u^2  + \frac32u^4 + \frac1{16}u^8 + \frac4{24}u^4 + \frac{1}{8} u^8 + \frac{54}{24^2}u^8 $$
As long as $u^2 < \frac12$, this gives
$$ \cos^4 u < 1 - 2u^2 + \frac34u^2 + \frac1{128}u^2 + \frac1{12}u^2 + \frac{1}{64}u^2 + \frac{54}{4068}u^2 $$
But $\frac34+\frac1{128}+\frac1{12}+\frac1{64}+\frac{54}{4068} < \frac{15}{20} + \frac1{20} + \frac2{20} + \frac1{20} + \frac1{20} = 1$, so this gives
$$ \cos^4 u < 1 - u^2 $$
as required.
A: Let $f(x):=\cos x-\cos^2(\sin x)$. It is sufficient to prove that $f(x)\gt 0$ for $0\lt x\lt \pi/4$.
Now,
$$f'(x)=-\sin x+2\cos(\sin x)\sin (\sin x)\cos x=-\sin x+\cos x\cdot \sin(2\sin x)$$$$=\cos x(\sin(2\sin x)-\tan x)$$
Here, let $g(x):=\sin(2\sin x)-\tan x$. Then,
$$g'(x)=\cos(2\sin x)\cdot 2\cos x-\frac{1}{\cos^2x}=\frac{2\cos^3x\cdot \cos(2\sin x)-1}{\cos^2x}$$
Here, let $h(x):=2\cos^3x\cdot \cos(2\sin x)-1$. Then,
$$h'(x)=6\cos^2x(-\sin x)\cdot \cos(2\sin x)+2\cos^3x(-\sin (2\sin x))\cdot 2\cos x\lt 0$$
So, $h(x)$ is decreasing with $h(0)=1,h(\pi/4)=\frac{1}{\sqrt 2}\cos(\sqrt 2)-1\lt 0$. So, there exists only one real $0\lt \alpha\lt\pi/4$ such that $h(\alpha)=0$. From this, $g'(x)$ is positive for $0\lt x\lt\alpha$, and is negative for $\alpha\lt x\lt\pi/4$. So, $g(x)$ is increasing for $0\lt x\lt\alpha$, and is decreasing for $\alpha\lt x\lt \pi/4$ with $g(0)=0,g(\pi/4)=\sin(\sqrt 2)-1\lt 0$. From this, there exists only one real $0\lt\beta\lt\pi/4$ such that $g(\beta)=0$. So, $f'(x)$ is positive for $0\lt x\lt\beta$, and is negative for $\beta\lt x\lt\pi/4$. So, $f(x)$ is increasing for $0\lt x\lt \beta$, and is decreasing for $\beta\lt x\lt\pi/4$ with $f(0)=0,f(\pi/4)=1/\sqrt 2-\cos^2(1/\sqrt 2)\gt 0$.
It follows from this that $f(x)\gt 0$ for $0\lt x\lt\pi/4$. The claim follows from this.
A: \begin{align*}
\sqrt{\cos x} & > \cos \sin x \\
\cos x & > \cos^2 \sin x \\
1 - 2 \sin^2 \frac x2 & > 1 - \sin^2 \sin x \\
\sin^2 \sin x & > 2 \sin^2 \frac x2 \\
\sin \sin x & > \sqrt 2 \sin \frac x2 \\
\end{align*}
Note that for $0<x<\frac \pi 4$ we have
$$\frac \pi 2 > \frac 1{\sqrt 2} = \sin \frac \pi 4 > \sin x = 2\sin \frac x2 \cos \frac x2 > 2 \sin \frac x2 \cos \frac \pi 8 > 0.$$
Since sine is increasing on the interval $\left(0,\frac\pi2\right)$ we can write
$$\sin \sin x > \sin \left(2\sin \frac x2 \cos \frac \pi 8 \right).$$
Therefore it is enough to prove that
$$\sin \left(2\sin \frac x2 \cos \frac \pi 8 \right) > \sqrt 2 \sin \frac x2.$$
Recall that for $t>0$ we have $\sin t > t - \frac{t^3}6$. Putting $t=2\sin \frac x2 \cos \frac \pi 8$ and keeping in mind that $2\sin \frac x2 \cos \frac \pi 8 < \frac 1{\sqrt 2}$ we obtain
\begin{align*}
\sin \left(2\sin \frac x2 \cos \frac \pi 8 \right) &> 2\sin \frac x2 \cos \frac \pi 8 - \frac 16 \left(2\sin \frac x2 \cos \frac \pi 8 \right)^3 >\\
&>2\sin \frac x2 \cos \frac \pi 8 - \frac 16 \left(\frac 1{\sqrt 2}\right)^2\left(2\sin \frac x2 \cos \frac \pi 8 \right) = \\
& = \frac{11}6 \sin \frac x2 \cos \frac \pi 8.
\end{align*}
Therefore it is enough to show that
$$\frac{11}6 \sin \frac x2 \cos \frac \pi 8 > \sqrt 2 \sin \frac x2$$
or
$$\frac{11}6 \cos \frac \pi 8 > \sqrt 2.$$
This is true as 
\begin{align*}
\frac{11}6 \cos \frac \pi 8 & > \frac 53 \cos \frac \pi 6 = \\
& = \frac 53 \cdot \frac{\sqrt 3}2 = \\
& = \sqrt{\frac{25}{12}} >\\
& > \sqrt 2.
\end{align*}
A: The only way I can think about is Taylor expansions (tedious but doable) as Henning Makholm commented. $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ $$\sqrt{\cos(x)}=1-\frac{x^2}{4}-\frac{x^4}{96}-\frac{19 x^6}{5760}+O\left(x^8\right)$$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O\left(x^8\right)$$ $$\cos(\sin(x))=1-\frac{x^2}{2}+\frac{5 x^4}{24}-\frac{37 x^6}{720}+\frac{457
   x^8}{40320}+O\left(x^9\right)$$ $$\sqrt{\cos(x)}-\cos(\sin(x))=\frac{x^2}{4}-\frac{7 x^4}{32}+\frac{277 x^6}{5760}+O\left(x^8\right)$$ $$\sqrt{\cos(x)}-\cos(\sin(x))=\frac{x^2}4 \left(1-\frac{7 x^2}{8}+ \frac{277 x^4}{1440}\right)+O\left(x^8\right)$$ There is no real root for the quadratic in $x^2$ inside brackets.
