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.

$f(x)=\frac{1-\cos x}{x^2}$ is an even function, so we only look at $x \in [0,1]$. It's easy to prove $f(x)$ is continuous and differentiable at $x=0$, and
$$f'(x) = \frac{x \sin x + 2 \cos x - 2}{x^3}$$
Now $$x \sin x + 2 \cos x - 2 \le 2 \tan \frac x2 \cdot \sin x - 4 \sin^2 \frac x2\\= 2 \tan \frac x2 \cdot 2 \sin \frac x2 \cos \frac x2 - 4 \sin^2 \frac x2=0, \forall x \in [0,1]$$
Hence $f(x) \ge f(1) = 1 - \cos 1\approx .46 > \frac 13, 1-\cos x \ge (1-\cos 1) x^2 > \frac 13 x^2.\blacksquare$
