Prove that $1-\frac{\sin(x)}{x} < x^2/2$ for $x \in (0, \pi/2)$

$1-\frac {\sin x} x \leq 1-\cos x=2\sin^{2} (\frac x 2)\leq 2(\frac x 2)^{2}=\frac {x^{2}} 2$.

Proof:

$1 - \frac{\sin(x)}{x} < 1 - \cos(x) = 2 (\sin(\frac{x}{2}))^2 < 2 . \frac{x}{2} . \frac{x}{2} $

Is this correct? I think that's the trick.

We can make the inequality slightly stronger. Taylor's theorem for $\sin(x)$ gives you $$1-\frac{\sin(x)}{x}=\cos(a)\frac{x^2}{6}$$ for some $a\in(0,x)$ and $$1-\frac{\sin(x)}{x}=\cos(a)\frac{x^2}{6}\leq\frac{x^2}{6} $$ which is in fact true for all $x\in\mathbb{R}$.