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