24th Korean Mathematical Olympiad Final Round

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 $$\frac{7}{18}$$