<s>[INST] Finding the maximum value of $\sum_{cyc}\frac{1}{a^2-4a+9}$ s.t. $a+b+c =1$ Problem: Let $a$, $b$, and $c$ be non-negative real numbers such that $a+b+c =1$. Find the maximum value of
\begin{align}
\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}.
\end{align}
By using the Karush–Kuhn–Tucker conditions, I know the maximum occurs at $a=0$, $b=0$, and $c=1$. But I want to know how to solve this problem without using calculus, because it is a math olympiad problem.
 [/INST] Another way:
By your work we need to prove that:
$$\sum_{cyc}\frac{1}{a^2-4a+9}\leq\frac{7}{18}$$ or
$$\sum_{cyc}\left(\frac{1}{a^2-4a+9}-\frac{1}{5}\right)\leq\frac{7}{18}-\frac{3}{5}$$ or
$$\sum_{cyc}\frac{(2-a)^2}{a^2-4a+9}\geq\frac{19}{18}.$$
Now, by C-S $$\sum_{cyc}\frac{(2-a)^2}{a^2-4a+9}=\sum_{cyc}\frac{(2-a)^2(4-a)^2}{(a^2-4a+9)(4-a)^2}\geq\frac{\left(\sum\limits_{cyc}(2-a)(4-a)\right)^2}{\sum\limits_{cyc}(a^2-4a+9)(4-a)^2}.$$
Id est, it's enough to prove that
$$18\left(\sum\limits_{cyc}(2-a)(4-a)\right)^2\geq19\sum\limits_{cyc}(a^2-4a+9)(4-a)^2,$$
which after homogenization gives:
$$\sum_{sym}(190a^3b+207a^2b^2+813a^2bc)\geq0,$$ which is obvious.
</s>