Question 92
Positive real $a$, $b$, $c$ satisfy the condition
$$
\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 1 + \frac{1}{6} \left( \frac{a}{c} + \frac{b}{a} + \frac{c}{b} \right).
$$
Prove that
$$
\frac{a^3bc}{b+c} + \frac{b^3ca}{c+a} + \frac{c^3ab}{a+b} \ge \frac{1}{6}(ab + bc + ca)^2.
$$
(I. Voronovich)
$$
\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 1 + \frac{1}{6} \left( \frac{a}{c} + \frac{b}{a} + \frac{c}{b} \right).
$$
Prove that
$$
\frac{a^3bc}{b+c} + \frac{b^3ca}{c+a} + \frac{c^3ab}{a+b} \ge \frac{1}{6}(ab + bc + ca)^2.
$$
(I. Voronovich)