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.

I shall give you the Hint:

*

*Assume that $f(x)=x^2\cdot \cos^2x$, and then prove, $$f'(x)=2x\cos x(\cos x-x\sin x)$$


*Find the maxima of $f(x)$ between $\left[0,{\pi\over2}\right]$. As $f(0)=f({\pi\over2})=0$ so there must be peak between $\left[0,{\pi\over2}\right]$. Hence, $f'(x)=0$ has the maxima. And for your inequality to be true maxima$<2$
