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.
 A: If $x<0$ then
$$\sin x<-x\implies 2x\sin x>-2x^2$$
and
$$\cos x\ge-1\implies x^2\cos x\ge-x^2$$
and so
$$3x^2+2x\sin x+x^2\cos x>3x^2-2x^2-x^2=0\ .$$
The case $x>0$ is   similar, even a bit easier. 
A: We have $3+\cos(x) \geq 2$ thus
$$3x^2+x^2\cos(x)>2x^2$$
Therefore
$$3x^2+2x\sin(x) + x^2\cos(x) > 2x(x+\sin(x)) \,.$$
Now, use the fact that both $2x$ and $x+\sin(x)$ are increasing and $0$ at $x=0$.
A: 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!
A: Another approach:
$$
3x^2+2xsin(x)+x^2cos(x) = x(3x+\sqrt{2+x^2}sin(\alpha+x))
$$
where, $\alpha = tan^{-1}(x/2)$
Since, $ x> 0$, we need only prove: $3x-\sqrt{2+x^2}>0$, for $x>0$
or $3x>\sqrt{2+x^2}$, squaring both sides, and rearranging terms, $x>\frac{1}{2}$
Also, for $0<x<\frac{1}{2}<\frac{\pi}{2}$, all terms in the expression are positive. 


*

* When $x>\frac{1}{2}>0$ the value as proved from the reduced expression is positive 

* When $0<x<\frac{1}{2}$, all values are positive and so the expression is positive.

