<s>[INST] How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.
 [/INST] For $|x| \le 1$ the Taylor series
$$
 \cos(x) = 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots
$$
is an alternating series with terms that decrease in absolute value. It follows that for these $x$
$$
 \cos(x) \le 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} \le  1 - \frac{x^2}{2!}+ \frac{x^2}{4!} = 1 - \frac{11}{24} x^2 
$$
and therefore
$$
 1 - \cos(x) \ge \frac{11}{24} x^2  \, .
$$
That is an even better estimate since $\frac{11}{24}  > \frac 13$.

The same approach can be used to show that $1 - \cos(x) \ge \frac{x^2}{3}$ holds on the larger interval $[-2, 2]$:
$$
 \cos(x) \le 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} \le 1 - \frac{x^2}{2!}+ \frac{4 x^2}{4!} = 1 - \frac 13 x^2 
$$
because the $x^{2n}/({2n})!$ terms decrease in absolute value for $n \ge 1$.
</s>