Let $2s = a + b + c$. Using $\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$, we obtain
$$\begin{aligned} 2 \sin^2 \left( \frac{C}{2} \right) &= 1 - \cos(C) = \frac{2ab - a^2 - b^2 + c^2}{2ab} \\ &= \frac{c^2 - (a-b)^2}{2ab} = \frac{(c-a+b)(c+a-b)}{2ab} \\ &= \frac{2(s-a)(s-b)}{ab} \end{aligned}$$

$$\sin\left(\frac{C}{2}\right) = \sqrt{\frac{(s-a)(s-b)}{ab}}$$
It follows, with the aid of the AM-GM inequality, that
$$\begin{aligned} 2ab \sin\left(\frac{C}{2}\right) &= 2\sqrt{(a(s-a))(b(s-b))} \\ &\le a(s-a) + b(s-b) = s(a+b) - a^2 - b^2, \end{aligned}$$
with equality iff $a(s - a) = b(s - b)$, i.e., $s(a - b) = (a - b)(a + b)$, i.e., iff $a = b$. Following from this we see, using $2ab \le a^2 + b^2$ etc., that
$$\begin{aligned} 2ab \sin \left(\frac{C}{2}\right) + 2bc \sin \left(\frac{A}{2}\right) + 2ca \sin \left(\frac{B}{2}\right) &\le 2s(a+b+c) - 2(a^2 + b^2 + c^2) \\ &= (a+b+c)^2 - 2(a^2 + b^2 + c^2) \\ &= 2ab + 2bc + 2ac - (a^2 + b^2 + c^2) \\ &\le (a^2 + b^2) + (b^2 + c^2) + (a^2 + c^2) - (a^2 + b^2 + c^2) \\ &= a^2 + b^2 + c^2, \end{aligned}$$
with equality iff $a = b = c$.