Prove that $\frac{x^2}{2} > x\cos x - \sin x$ for every $x\neq0$ I'm trying to prove that $\frac{x^2}{2} > x\cos x - \sin x$ for every $x\neq0$
$f(x) =\frac{x^2}{2} - x\cos x + \sin x $.
I need to prove that $f > 0 $ for every $x\neq 0$.
$f(0) = 0$.
$f'(x) = x(1+\sin x)$
for every $x > 0$,
$ f'(x) \geq 0$ and this means that $f$ is strictly growing in that interval meaning that for $x>0, f(x)>0$
However for negative values $f'(x)$ is not always positive. If so, how can I prove that $f >0 $ for those values as well?
 A: We have that $\frac{x\cos x-\sin x}{x^2}=\frac{d}{dx}\left(\frac{\sin x}{x}\right)$ is an odd function, hence in order to prove is is bounded by $\frac{1}{2}$ for any $x\in\mathbb{R}$ it is enough to show that its absolute value is bounded by $\frac{1}{2}$ for any $x\geq 0$. 
If $x\geq 0$ we have $\sin(x)\leq\min(x,1)$, hence
$$ \forall t\geq 0,\qquad t\sin(t) \leq \min(t,t^2)\tag{1} $$
and by integrating both sides of $(1)$ over the interval $(0,x)$ we get:
$$ \forall x\geq 1,\qquad \sin(x)-x\cos(x)\leq \frac{x^2}{2}-\frac{1}{6} \tag{2}$$
$$ \forall x\in[0,1],\qquad \sin(x)-x\cos(x)\leq \frac{x^3}{6}\tag{3} $$
It follows that over the interval $\left[0,\sqrt{\frac{7}{3}+2\sqrt{\frac{7}{3}}}\right]$ the function $f(x)=\frac{\sin x-x\cos x}{x^2} $ is positive and bounded by $\frac{1}{5}\left(3-\sqrt{\frac{3}{7}}\right)$. On the other hand, by the Cauchy-Schwarz inequality we have $\left|x\cos(x)-\sin(x)\right|\leq \sqrt{x^2+1}$, hence
$$ \left|\frac{x\cos(x)-\sin(x)}{x^2}\right|\leq \frac{1}{5}\left(3-\sqrt{\frac{3}{7}}\right) \tag{4} $$
is trivial for any $x\geq \sqrt{\frac{7}{3}+2\sqrt{\frac{7}{3}}}$. This proves the slightly stronger inequality

$$ \forall x\in\mathbb{R},\qquad \frac{x\cos(x)-\sin(x)}{x^2}\leq \color{red}{\frac{1}{5}\left(3-\sqrt{\frac{3}{7}}\right)}=0.4690692658584\ldots\tag{5}$$

