Mongolian Mathematical Olympiad

First, we will prove if $\forall x,y>0$ then the inequality $(\ast )\frac{x^2-xy+y^2}{x^2+xy+y^2}\ge \frac{1}{3}$ holds.

$(\ast )\iff 3(x^2-xy+y^2)\ge x^2+xy+y^2\iff 2(x^2+y^2)-4xy\ge 0\iff 2(x-y{)}^2\ge 0$ thus proof of $(\ast )$ completes. Consequently, $$\frac{a^2(a^3+b^3)}{a^2+ab+b^2}=a^2(a+b)\frac{a^2-ab+b^2}{a^2+ab+b^2}\ge \frac{1}{3}{a}^2(a+b)\text{ by }(\ast ).$$ Similarly, $\frac{b^2(b^3+c^3)}{b^2+bc+c^2}\ge \frac{1}{3}{b}^2(b+c)$ and $\frac{c^2(c^3+a^3)}{c^2+ca+a^2}\ge \frac{1}{3}{c}^2(c+a)$ from where follows $$\frac{a^2(a^3+b^3)}{a^2+ab+b^2}+\frac{b^2(b^3+c^3)}{b^2+bc+c^2}+\frac{c^2(c^3+a^3)}{c^2+ca+a^2}\ge \frac{1}{3}(a^3+b^3+c^3)+\frac{1}{3}(a^{2}b+b^{2}c+c^{2}a)\ge \frac{1}{3}\cdot 3\sqrt{a^3{b}^3{c}^3}+\frac{1}{3}\cdot 3\sqrt{a^{2}b\cdot b^{2}c\cdot c^{2}a}=2abc\text{ and we have done. Equality holds when }a=b=c.$$