<s>[INST] prove $\forall x \in (0, \frac{\pi}{2})$: $x^2 \cos^2 (x) < 2$ Prove  $\forall x \in (0, \frac{\pi}{2})$: $x^2 \cos^2 (x) < 2$
It is clear, that:

*

*$x^2 \cos^2 (x) \to 0, x \to \frac{\pi}{2}$, since cosine goes to 0.

*$x^2 \cos^2 (x) \to 0, x \to 0$, since $x$ goes to 0.

Of course, we can split $(0, \frac{\pi}{2})$ into $A_1 = (0, \frac{\pi}{4}]$, $ A_2 = [\frac{\pi}{4}, \frac{\pi}{3}]$ and $A_3 = [\frac{\pi}{3}, \frac{\pi}{2})$
(sorry for intersection, just don't want to write limits. It doesn't affect anything)
Then we take max and min values from $A_i$ and evaluate max value at $x^2$ and min value at $\cos^2 x$:
$A_1$: $\max(A_1)^2 = \frac{\pi^2}{16} \in (0, 1); \cos^2 \min(A_1) = 1 \implies \max(A_1)^2 \cos^2 \min(A_1) < 2$
$A_2$: $\max(A_2)^2 = \frac{\pi^2}{9} \in (1, 2) ; \cos^2 \min(A_2) = \frac{1}{2} \implies 
 \max(A_2)^2 \cos^2 \min(A_2) < 2$
$A_3$: $\max(A_3)^2 = \frac{\pi^2}{4} \in (2, 3) ; \cos^2 \min(A_3) = \frac{1}{4} \implies 
 \max(A_3)^2 \cos^2 \min(A_3) < 2$
Actually, here we can see that upper bound is even lower (not bigger that 1).
Unfortunately, I am not satisfied with this writing (somehow I am sure there are much better proofs).
But what are they? Thank you.
 [/INST] This is based on the inequality $x< \tan x$ for $x\in\left(0, \frac \pi2\right)$.
(Geometrically, this can be obtained by comparing the areas of 1) a sector with unit radius and angle $x$, and 2) a right-angled triangle with adjacent side $1$ and opposite side $\tan x$.)
$$\begin{align*}
x &< \tan x\\
x^2 \cos^2 x &< \tan^2x\cos ^2x\\
&= \sin^2 x < 1 < 2
\end{align*}$$
</s>