<s>[INST] either inequality $\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$ is hard or I need to go back to study ASAP Prove that for every $x \in(0,\frac{\pi}{2})$, the following inequality:
$\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$
holds
I don't see room to use derivatives, since it seems a little messy to calculate the $\lim_{x\to 0}$ of $\frac{2\ln(\cos{x})}{x^2}$
 (which, I think, is necessary in order to make usage of derivatives).
Any hints? I've already tried cross multiplying but it doesn't lead anywhere, unless I missed something.
 [/INST] You need the following two inequalities:
 $$\cos x = 1- \frac{x^2}{2}+ \frac{x^4}{24}-\frac{x^6}{6!} + \cdots < 1- \frac{x^2}{2}+ \frac{x^4}{24}$$
and
$$\ln (1+y) = y -\frac{y^2}{2}+ \cdots\le y$$
(you can formally prove them using derivatives, if you want).
Combining them substituting $y= - \frac{x^2}{2}+ \frac{x^4}{24}$ you get
$$\ln \cos x < \ln \left(1- \frac{x^2}{2}+ \frac{x^4}{24} \right) \le - \frac{x^2}{2}+ \frac{x^4}{24}$$
Multiply both sides by $\frac{2}{x^2}$ to get
$$\frac{2 \ln ( \cos x)}{x^2} < -1+\frac{x^2}{12}$$
</s>