Prove that $-x\sin x<\frac{x^4}{6}-x^2$ for all $x\in(0,\pi)$ Prove that $-x\sin x<\frac{x^4}{6}-x^2$ for all $x\in(0,\pi)$.
My approach is that for similar type of problem I used two function
$f(x)=y=-x\sin x$ & $g(x)=y=\frac{x^4}{6}-x^2$ 
$f(x)-g(x)<0$ for $x\in(0,\pi)$
After this step I am confused.

Recall Maclaurin's series of $\sin(x)$, that is
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...+\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$
Then
$$-x\sin x=-x^2+\frac{x^4}{6}-\frac{x^6}{120}+...+\frac{(-1)^{n+1}}{(2n+1)!}x^{2n+2}$$
We rewrite above Maclaurin expansion of $-x \sin x$ as
$$-x\sin x=-x^2+\frac{x^4}{6}-\frac{x^6}{120}+O\left(\frac{x^6}{120}\right)$$
Therefore, for every $x\in(0, \pi)$ inequality holds
$$-x^2+\frac{x^4}{6}-\frac{x^6}{120}+O\left(\frac{x^6}{120}\right)< -x^2+\frac{x^4}{6}$$
This completes the proof.
