<s>[INST] 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.
 [/INST] 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.

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