<s>[INST] 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...
 [/INST] 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.
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