China Western Mathematical Olympiad

$$\text{If } a < \frac{9}{5}, \text{ then}$$
$$\frac{1}{5a^2 - 4a + 11} \le \frac{1}{24}(3-a). \quad \textcircled{1}$$
In fact,
$$\begin{align*} \textcircled{1} &\Leftrightarrow (3-a)(5a^2-4a+11) \ge 24 \\ &\Leftrightarrow 5a^3 - 19a^2 + 23a - 9 \le 0 \\ &\Leftrightarrow (a-1)^2(5a-9) \le 0 \Leftrightarrow a < \frac{9}{5}. \end{align*}$$
So if $a$, $b$, $c < \frac{9}{5}$, then
$$\begin{align*} & \frac{1}{5a^2 - 4a + 11} + \frac{1}{5b^2 - 4b + 11} + \frac{1}{5c^2 - 4c + 11} \\ &\le \frac{1}{24}(3-a) + \frac{1}{24}(3-b) + \frac{1}{24}(3-c) \\ &= \frac{1}{4}. \end{align*}$$
If one of $a$, $b$, $c$ is not less than $\frac{9}{5}$, say $a \ge \frac{9}{5}$, then
$$\begin{aligned} 5a^2 - 4a + 11 &= 5a \left(a - \frac{4}{5}\right) + 11 \\ &\ge 5 \cdot \frac{9}{5} \cdot \left(\frac{9}{5} - \frac{4}{5}\right) + 11 = 20. \end{aligned}$$
$$\text{So} \quad \frac{1}{5a^2 - 4a + 11} \le \frac{1}{20}.$$
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
$$\begin{aligned} & \frac{1}{5a^2 - 4a + 11} + \frac{1}{5b^2 - 4b + 11} + \frac{1}{5c^2 - 4c + 11} \\ < & \frac{1}{20} + \frac{1}{10} + \frac{1}{10} = \frac{1}{4}. \end{aligned}$$
Hence the inequality holds for all $a$, $b$, $c$.