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...

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.
