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...

\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*}
