<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] Rewriting the inequality $$2\sin^2(\frac{x}{2})\ge\frac{x^2}3 \iff (\sqrt2 \sin(\frac{x}{2})  - \frac{x}{\sqrt3})(\sqrt2 \sin(\frac{x}{2})  + \frac{x}{\sqrt3}) \ge 0 \tag{1}$$
Assume $0\le x \le 1$. Obviously we have $$\sqrt2 \sin(\frac{x}{2})  + \frac{x}{\sqrt3}\ge 0$$ So we should analyze $$f(x,a) = \sin(\frac{x}{2}) - \frac{x}{a}$$It's interesting to see this visually: 
When $a=2$, the line $y_1 = \frac{x}{2}$ touches $y_2 = \sin(\frac{x}{2})$ only at $x = 0$(remember that $|\sin(\frac{x}{2})|\le |\frac{x}{2}|$). Increasing $a$ further, leads to another intersection point(we can actually see the concavity of $\sin(\frac{x}{2})$ here).
Note that $$f(x,a) = \int_{0}^{x}(\frac{1}{2}\cos (\frac{t}{2}) - \frac{1}{a})dt$$Let $a = \sqrt6$. We can show that the integrand is nonnegative when $0\le t \le 1$. So when $0\le x \le 1$ we have $f(x,\sqrt6)\ge 0 $ which implies $$\sqrt2 \sin(\frac{x}{2})  - \frac{x}{\sqrt3}\ge 0$$In order to show nonnegativity, consider $$0\le t \le 1 \implies 0\le \frac{t}{2} \le \frac{1}{2} \lt \frac{\pi}{2} \implies 1 \ge \cos(\frac{t}{2}) \ge \cos(\frac{1}{2}) \implies \frac{1}{2} \ge \frac{1}{2}\cos(\frac{t}{2}) \ge \frac{1}{2}\cos(\frac{1}{2}) \approx 0.439 \gt \frac{1}{\sqrt6} \approx 0.408$$ Now that we have shown $(1)$ holds for $0 \le x \le 1$, let $x = -t$ and conclude it holds for $-1 \le t \le 0$.
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