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