TITLE: How to prove non negativeness of $f(x)=1-\cos x-\frac{x^2}{20}$
QUESTION [5 upvotes]: 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?

REPLY [5 votes]: $$1-\cos x=2\sin^2
\frac x2.$$
We need
$$\sin^2 \frac x2\ge\frac{x^2}{10}$$
for $|x|\le\pi$, equivalently
$$\sin^2y\ge\frac{2y^2}5$$
for $|y|\le\frac\pi2$. Again this is equivalent to
$$\sin y\ge\frac{y}{\sqrt{5/2}}\tag1$$
for $0\le y\le\frac\pi2$.
But it is well-known that
$$\sin y\ge\frac{y}{\pi/2}\tag2$$
for $0\le y\le\frac\pi2$ (basically convexity). Then (1) follow from (2)
provided that $\pi/2\le\sqrt{5/2}$, equivalently $\pi^2\le10$.
You can check this on your pocket calculator.