Selection and Training Session

By the Cauchy-Bunyakovskii inequality, we have $$(a^{2}b+b^{2}c+c^{2}a)(a+b+c)\ge (ab+bc+ca{)}^2.(\ast )$$

Now consider the given equality. Multiplying both its sides by $(a+b+c)abc$, we get $$A=(a+b+c)\left(\frac{a^{2}bc}{b+c}+\frac{b^{2}ca}{c+a}+\frac{c^{2}ab}{a+b}\right)=abc(a+b+c)+\frac{(a+b+c)(a^{2}b+b^{2}c+c^{2}a)}{6}\ge \text{[see }(\ast )\text{]}\ge abc(a+b+c)+\frac{1}{6}(ab+bc+ca{)}^2.$$ Further, $$A=\sum _{cyc}\left(\frac{a^{3}bc}{b+c}+a^{2}bc\right)=\sum _{cyc}\frac{a^{3}bc}{b+c}+abc(a+b+c)\ge abc(a+b+c)=\frac{1}{6}(ab+bc+ca{)}^2,$$ whence the required inequality follows.