Showing that $3x^2+2x\sin(x) + x^2\cos(x) > 0$ for all $x\neq 0$ I got this question:
Show that for all $x\neq 0$, $3x^2+2x\sin(x) + x^2\cos(x) > 0$
I tried to show it but got stuck.

If $x > 0$, then: $f(x) = x(3x + 2\sin x + x\cos x)$. We need to prove:
$3x + 2\sin x + x\cos x > 0$ when $x > 0$.
$3x + 2\sin x + x\cos x \geq 3x + 2\sin x - x = 2(x + \sin x) > 0$ because $x + \sin x > 0$ as $(x + \sin x)' = 1 + \cos x \geq 0$. So: $x + \sin x > 0 + \sin0 = 0$.
If $x < 0$ we again need to show: $3x + 2\sin x + x\cos x < 0$. But:
$3x + 2\sin x + x\cos x \leq 3x + 2\sin x - x = 2(x + \sin x) < 0$ by the same argument as above. Done!
