, 2010

Algebra Difficulty 4.9 AIME Prove it Ukraine

For what real values $a$ and $b$ , maximum among $3a^2 + 2b$ and $3b^2 + 2a$ takes minimum value?

Solution

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 $a=b=-\frac{1}{3}$ .

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