Mathematical Reflections 2012, Issue 2 - Problem J225

Problem:
Let $a, b, c$ be nonnegative real numbers such that $a+b+c=1$. Prove that $$ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)+abc \leq \dfrac{1}{8}.$$

Proposed by Titu Andreescu.

Solution:

Using the fact that $abc=abc(a+b+c)$ and factorizing we have
$$ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)+abc=(a^2+b^2+c^2)(ab+bc+ca).$$
By AM-GM Inequality, $$a^2+b^2+c^2+2(ab+bc+ca)=1 \geq 2\sqrt{2(a^2+b^2+c^2)(ab+bc+ca)},$$
then $$(a^2+b^2+c^2)(ab+bc+ca) \leq \dfrac{1}{8}.$$