Let $ f(x) = \frac {1}{5x^2 - 4x + 11}.$ Then $ f'(x) = \frac {4 - 10x}{(5x^2 - 4x + 11)^2}.$ Hence, $ f(x)\leq f\left(\frac {2}{5}\right)$ for all real $ x$ and $ f$ is a decreasing function on $ [\frac {2}{5}, + \infty].$ But $ \sum_{cyc}f(a)\leq\frac {1}{4}\Leftrightarrow\sum_{cyc}\left(\frac {1}{12} - \frac {a - 1}{24} - \frac {1}{5a^2 - 4a + 11}\right)\geq0\Leftrightarrow$ $ \Leftrightarrow\sum_{cyc}\frac {(9 - 5a)(a - 1)^2}{5a^2 - 4a + 11}\geq0.$ If $ \max\{a,b,c\}\leq\frac {9}{5}$ then our inequality is true. If $ \max\{a,b,c\} > \frac {9}{5}$ then $ \sum_{cyc}f(a) < f\left(\frac {9}{5}\right) + 2f\left(\frac {2}{5}\right) < \frac {1}{4}.$

It is the official solution,very nice!Thank you.

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A stronger inequality is : if $ a,b,c$ be real numbers such that $ a + b + c = 3$, then $ \frac {1}{9a^2 - 2a + 25} + \frac {1}{9b^2 - 2b + 25} + \frac {1}{9c^2 - 2c + 25}\leq\frac {3}{32},$ with equality if $ a=\frac{17}{9},b=c=\frac{5}{9}.$ See also here : http://www.mathlinks.ro/viewtopic.php?t=223910

Ji Chen wrote: A stronger inequality is : if $ a,b,c$ be real numbers such that $ a + b + c = 3$, then $ \frac {1}{9a^2 - 2a + 25} + \frac {1}{9b^2 - 2b + 25} + \frac {1}{9c^2 - 2c + 25}\leq\frac {3}{32},$ with equality if $ a = \frac {17}{9},b = c = \frac {5}{9}.$ See also here : http://www.mathlinks.ro/viewtopic.php?t=223910 Denote $ q=ab+bc+ca$ and $ r=abc$ then \[ \frac {1}{{9a^2 - 2a + 25}} + \frac {1}{{9b^2 - 2b + 21}} + \frac {1}{{9c^2 - 2c + 21}} \le \frac {3}{{32}} \] \[ \Leftrightarrow f\left( r \right) = 2187r^2 - \left( {486q + 18276} \right)r + 3483q^2 - 7100q + 7500 \ge 0 \] We have $ f'\left( r \right) = 4374r - 486q - 18276 \le 0$ thus it suffices us to prove the inequality in the case $ a = b$. Hence the inequality becomes \[ \frac {2}{{9a^2 - 2a + 25}} + \frac {1}{{9\left( {3 - 2a} \right)^2 - 2\left( {3 - 2a} \right) + 21}} \le \frac {3}{{32}} \] \[ \Leftrightarrow 12\left( {a - 1} \right)^2 \left( {9a - 5} \right)^2 \ge 0 \] which is clearly true. Equality holds if and only if $ a = b = c = 1$ or $ a = b = \frac {5}{9}$ and $ c = \frac {{17}}{9}$ or any cyclic permutations, .

Ji Chen wrote: If $ a,b,c$ be real numbers such that $ a + b + c = 3$, then $ \frac {1}{9a^2 - 2a + 25} + \frac {1}{9b^2 - 2b + 25} + \frac {1}{9c^2 - 2c + 25}\leq\frac {3}{32}.$ $ \frac {3}{32} - \sum{\frac {(a + b + c)^2}{81a^2 - 6a(a + b + c) + 25(a + b + c)^2}}$ $ \equiv\frac {3F(a,b,c)}{8\prod{\left[81a^2 - 6a(a + b + c) + 25(a + b + c)^2\right]}},$ $ F(a,b,c) = F(a,s + t,s - t)$ $ = 2187(43s^2 + 34sa + 67a^2)t^4$ $ - 6(17147s^4 + 4252s^3a + 1434s^2a^2 - 35828sa^3 - 7741a^4)t^2$ $ + (s - a)^2(17s - 5a)^2(169s^2 + 94sa + 25a^2)\geq0.$

The following inequality is also true. Let $ a,b,c$ be real numbers such that $ a+b+c=3$. Prove that \[\frac{a}{5a^2-4a+11}+\frac{b}{5b^2-4b+11}+\frac{c}{5c^2-4c+11}\leq\frac{1}{4}.\]

sqing wrote: The following inequality is also true. Let $ a,b,c$ be real numbers such that $ a+b+c=3$. Prove that \[\frac{a}{5a^2-4a+11}+\frac{b}{5b^2-4b+11}+\frac{c}{5c^2-4c+11}\leq\frac{1}{4}.\] Because $\frac{1}{4}-\sum_{cyc}\frac{a}{5a^2-4a+11}\geq\frac{5\left(\sum\limits_{cyc}(8a^3-15a^2b-15a^2c+22abc)\right)^2}{2916\prod\limits_{cyc}(5a^2-4a+11)}\geq0$.

Ji Chen wrote: A stronger inequality is : if $ a,b,c$ be real numbers such that $ a + b + c = 3$, then $ \frac {1}{9a^2 - 2a + 25} + \frac {1}{9b^2 - 2b + 25} + \frac {1}{9c^2 - 2c + 25}\leq\frac {3}{32},$ with equality if $ a=\frac{17}{9},b=c=\frac{5}{9}.$ See also here : http://www.mathlinks.ro/viewtopic.php?t=223910 A weaker inequality： $$ \frac {1}{9a^2 - 6a + 21} + \frac {1}{9b^2 - 6b + 21} + \frac {1}{9c^2 - 6c + 21}\leq\frac {1}{8}$$ sqing wrote: The following inequality is also true. Let $ a,b,c$ be real numbers such that $ a+b+c=3$. Prove that \[\frac{a}{5a^2-4a+11}+\frac{b}{5b^2-4b+11}+\frac{c}{5c^2-4c+11}\leq\frac{1}{4}.\] Let $ a,b,c$ be non-negative real numbers such that $ a+b+c=3$. Prove that \[\frac{3}{44}\leq\frac{a}{5a^2-4a+11}+\frac{b}{5b^2-4b+11}+\frac{c}{5c^2-4c+11}\leq\frac{1}{4}.\]