Competition problem — Algebra — Olympiad Maths Prep

Algebra Difficulty 5.2 AIME, harder Prove it South Korea

Find the maximum value of
$\frac{1}{a^2 - 4a + 9} + \frac{1}{b^2 - 4b + 9} + \frac{1}{c^2 - 4c + 9}$
where $a$ , $b$ , $c$ are non-negative real numbers satisfying $a + b + c = 1$ .

Solution

Note that for $0 \le x \le 1$ the inequality
$\frac{1}{x^2 - 4x + 9} \le \frac{x + 2}{18}$
holds, where the equality holds if and only if $x = 0$ or $x = 1$ . Hence,
$\frac{1}{a^2 - 4a + 9} + \frac{1}{b^2 - 4b + 9} + \frac{1}{c^2 - 4c + 9} \le \frac{1}{18}(a + b + c + 6) = \frac{7}{18}$
Since the equality holds for $a = 0$ , $b = 0$ , $c = 1$ , the maximum value we look for is
$\frac{7}{18}$

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