How to prove non negativeness of $f(x)=1-\cos x-\frac{x^2}{20}$ I'm trying to prove that $f(x)=1-\cos x -\frac{x^2}{20}$, defined on $[-\pi, \pi]$, is a non negative function.
How do I prove that $f(x)\ge 0 $ for all $x \in [-\pi, \pi]$?
Hints?

HINT
Since
$$1-\frac{x^2}{2}\le \cos(x)\le1-\frac{x^2}{2}+\frac{x^4}{24}$$
Thus
$$f(x)=1-\cos x -\frac{x^2}{20}\geq 1-1+\frac{x^2}{2}-\frac{x^4}{24}-\frac{x^2}{20}=\frac{9x^2}{20}-\frac{x^4}{24}\geq 0 \quad x\in[-\pi,\pi]$$
Indeed
$$\frac{9x^2}{20}-\frac{x^4}{24}=x^2\left(\frac{9}{20}-\frac{x^2}{24}\right)\geq0 \iff x^2 \leq \frac{216}{20}=\frac{54}{5} \iff \\x\in \left[-\sqrt{\frac{54}{5}},\sqrt{\frac{54}{5}}\right]\implies x\in\left[-\pi,\pi\right]$$
