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.

The Maclaurin series of the tangent function has all coefficients positive (see formula). Integrating we get that the Maclaurin series of the function $-\log(\cos x)$ has all coefficients positive. Up to order $12$ it is 
$$-\log (\cos x) =\frac{x^2}{2} + \frac{x^4}{12} + \frac{x^6}{45} + \frac{17 x^8}{2520} + \frac{31 x^{10}}{14175} + \frac{691 x^{12}}{935550} + 
\mathcal{O}(x^{14})$$
So the inequality should probably read 
$$-\frac{x^2}{2} -\frac{x^4}{12}-\log(\cos x) >0$$ or
$$ \frac{2 \log(\cos(x))}{x^2}<-1 - \frac{x^2}{6}$$
ADDED: 
We can prove the inequality $-\frac{x^2}{2} -\frac{x^4}{12}-\log(\cos x) >0$ by taking the derivatives of the function $-\frac{x^2}{2} -\frac{x^4}{12}-\log(\cos x) $ up to order $5$. One can check easily that they all take value $0$ at $0$. Then one calculates:
$$\left(-\frac{x^2}{2} -\frac{x^4}{12}-\log(\cos x) \right)^{(5)}=4(5-\cos 2x)\sec^4 x \tan x>0$$ for $0<x<\pi/2$. From here one obtains the inequality . Note that we also get $\tan x>x + \frac{x^3}{3}$. 
