50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010)

Let $M(a,b)=\max \{3a^2+2b,3b^2+2a\}$. Then $M(a,b)\ge 3a^2+2b$ and $M(a,b)\ge 3b^2+2a$. From the last two inequalities we get $2M(a,b)\ge 3a^2+2b+3b^2+2a$.

We now have $$\frac{2}{3}M(a,b)+\frac{2}{9}\ge {\left(a+\frac{1}{3}\right)}^2+{\left(b+\frac{1}{3}\right)}^2\ge 0,$$ or $M(a,b)\ge -\frac{1}{3}$ and $M(a,b)=-\frac{1}{3}$, if $a=b=-\frac{1}{3}$.