<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] Note that since $e^x$ is strictly increasing
$$\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1\iff\ln \cos x< \frac{x^4}{24}- \frac{x^2}{2}\iff \cos x<e^{\frac{x^4}{24}- \frac{x^2}{2}}$$
and since


*

*$e^x>1+x$

*$\cos x < 1-\frac{x^2}{2}+\frac{x^4}{24}$ 


(refer to Using Taylor's theorem show that $1 - \frac{x^2}{2} < \cos x < 1- \frac{x^2}{2} + \frac{x^4}{24}$)
we have
$$e^{\frac{x^4}{24}- \frac{x^2}{2}}>1-\frac{x^2}{2}+\frac{x^4}{24}>\cos x$$
</s>