Ukrainian Mathematical Olympiad

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

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 $$\begin{array}{rl}\frac{a+b}{c^2}+\frac{b+c}{a^2}+\frac{c+a}{b^2}& =\left(\frac{a}{b^2}+\frac{b}{a^2}\right)+\left(\frac{b}{c^2}+\frac{c}{b^2}\right)+\left(\frac{c}{a^2}+\frac{a}{c^2}\right)\\ & \ge \left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{c}+\frac{1}{a}\right)=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).\end{array}$$ That is, $$\frac{a+b}{c^2}+\frac{b+c}{a^2}+\frac{c+a}{b^2}\ge 2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).(2)$$ It remains to multiply inequalities (1) and (2).