It is easy to prove the inequalities
$$\frac{a^2}{b+c} \geq a - \frac{b+c}{4}, \quad \frac{b^2}{c+a} \geq b - \frac{c+a}{4}, \quad \frac{c^2}{a+b} \geq c - \frac{a+b}{4},$$
adding which we obtain
$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \ge \frac{a+b+c}{2}. \quad (1)$$

Since, as can be easily verified, $\frac{x}{y^2} + \frac{y}{x^2} \ge \frac{1}{x} + \frac{1}{y}$ for $x>0, y>0$, then
$$\begin{align*} \frac{a+b}{c^2} + \frac{b+c}{a^2} + \frac{c+a}{b^2} &= \left( \frac{a}{b^2} + \frac{b}{a^2} \right) + \left( \frac{b}{c^2} + \frac{c}{b^2} \right) + \left( \frac{c}{a^2} + \frac{a}{c^2} \right) \\ &\ge \left( \frac{1}{a} + \frac{1}{b} \right) + \left( \frac{1}{b} + \frac{1}{c} \right) + \left( \frac{1}{c} + \frac{1}{a} \right) = 2 \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right). \end{align*}$$
That is,
$$\frac{a+b}{c^2} + \frac{b+c}{a^2} + \frac{c+a}{b^2} \ge 2 \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right). \quad (2)$$
It remains to multiply inequalities (1) and (2).