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.

Fix $x\in[0,1]$. Using
$$ \sin(\frac12x)\ge \frac x2-\frac{1}{6}(\frac{x}{2})^3=\frac x2(1-\frac {x^2}{24})\ge \frac x2(1-\frac1{24})=\frac{23x}{48}  $$
you can have
\begin{eqnarray}
1-\cos x&=&2\sin^2(\frac x2)\ge2(\frac{23x}{48})^2>\frac{x^2}{3}.
\end{eqnarray}
