China Western Mathematical Olympiad

$$\text{If }a<\frac{9}{5},\text{ then}$$ $$\frac{1}{5a^2-4a+11}\le \frac{1}{24}(3-a).\stackrel{◯}{1}$$ In fact, $$\begin{array}{rl}\stackrel{◯}{1}& \iff (3-a)(5a^2-4a+11)\ge 24\\ & \iff 5a^3-19a^2+23a-9\le 0\\ & \iff (a-1{)}^2(5a-9)\le 0\iff a<\frac{9}{5}.\end{array}$$ So if $a$, $b$, $c<\frac{9}{5}$, then $$\begin{array}{rl}& \frac{1}{5a^2-4a+11}+\frac{1}{5b^2-4b+11}+\frac{1}{5c^2-4c+11}\\ & \le \frac{1}{24}(3-a)+\frac{1}{24}(3-b)+\frac{1}{24}(3-c)\\ & =\frac{1}{4}.\end{array}$$ If one of $a$, $b$, $c$ is not less than $\frac{9}{5}$, say $a\ge \frac{9}{5}$, then $$\begin{array}{rl}5a^2-4a+11& =5a\left(a-\frac{4}{5}\right)+11\\ & \ge 5\cdot \frac{9}{5}\cdot \left(\frac{9}{5}-\frac{4}{5}\right)+11=20.\end{array}$$ $$\text{So}\frac{1}{5a^2-4a+11}\le \frac{1}{20}.$$ Since $$5b^2-4b+11=5{\left(b-\frac{2}{5}\right)}^2+11-\frac{4}{5}\ge 11-\frac{4}{5}>10,$$ we have $\frac{1}{5b^2-4b+11}<\frac{1}{10}$. Similarly, $\frac{1}{5c^2-4c+11}<\frac{1}{10}$. So $$\begin{array}{rl}& \frac{1}{5a^2-4a+11}+\frac{1}{5b^2-4b+11}+\frac{1}{5c^2-4c+11}\\ <& \frac{1}{20}+\frac{1}{10}+\frac{1}{10}=\frac{1}{4}.\end{array}$$ Hence the inequality holds for all $a$, $b$, $c$.