IRL ABooklet

The half-angle formula, $2\sin^2(A/2) = 1 - \cos A$, and the Cosine-Rule give
$$4 \sin^2 \frac{A}{2} = 2(1 - \cos A) = 2 - \frac{b^2 + c^2 - a^2}{bc} = \frac{a^2 - (b-c)^2}{bc} \le \frac{a^2}{bc},$$
with equality iff $b = c$. Because $\sin(A/2) > 0$, we can take square roots to obtain the first inequality. The second and third inequalities follow immediately. Here is another approach to the third one:
$$\begin{aligned} & 4 \left( \sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} \right) \\ &= 2(1 - \cos A) + 2(1 - \cos B) + 2(1 - \cos C) \\ &= 6 - \frac{b^2 + c^2 - a^2}{bc} - \frac{c^2 + a^2 - b^2}{ca} - \frac{a^2 + b^2 - c^2}{ab} \\ &= 6 - \left( \frac{b}{c} + \frac{c}{b} + \frac{a}{c} + \frac{a}{c} + \frac{a}{b} + \frac{b}{a} \right) + \frac{a^3 + b^3 + c^3}{abc} \\ &= 9 - (a+b+c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) + \frac{a^3 + b^3 + c^3}{abc} \\ &\le \frac{a^3 + b^3 + c^3}{abc}, \end{aligned}$$
since, by the HM-AM inequality, for any positive numbers $x, y, z$,
$$(x + y + z) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \ge 9,$$
with equality iff $x = y = z$.