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.
 A: Recall that by Taylor's series we have
$$\sin x >x-\frac16 x^3 \implies -x\sin x<\frac16 x^4-x^2$$
for the proof refer to Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$
Following your way we need to show that
$$f(x)=\sin x - x +\frac16 x^2>0$$
then it suffices to note that $f(0)=0$ and that $f(x)$ is strictly ingreasing that is
$$f'(x)=\cos x -1+\frac12 x^2>0$$
for which it suffices to note that $f'(0)=0$ and that $f'(x)$ is strictly ingreasing that is
$$f''(x)=-\sin x+x>0\implies \sin x < x$$
which is true (we can prove also that by continuing the derivation process).
A: 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.
