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.

The positive root of the equation $1-\cos x = \frac{x^2}{3}$ is approximately $2.16$
so the inequality is valid on a larger interval, say $[-2, 2]$. To show this consider
$$\frac{1-\cos x}{\frac{x^2}{3}} = \frac{2 \sin^2 \frac{x}{2}}{\frac{x^2}{3}}= \frac{3}{2} \left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2$$
Now, the function $t\mapsto \sin t$ is concave on $[0, \pi]$, so $\frac{\sin t}{t}$ is decreasing on $[0, \pi]$. Therefore, on the interval $[-2, 2]$ we have $\frac{\sin \frac{x}{2}}{\frac{x}{2}} \ge \sin 1$ and so
$$\frac{1-\cos x}{x^2/3} \ge \frac{3}{2} \sin^2 1= 1.06.. > 1$$
On the interval $[-1,1]$ the inequality is
$$1- \cos x \ge 2 \sin^2\frac{1}{2}\cdot  x^2$$
where $2 \sin^2 \frac{1}{2} = 0.45969\ldots$
and this is the best estimate on $[-1,1]$.
