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.

Because$:$
$$\dfrac{a}{18} + \dfrac{1}{9} - \dfrac{1}{a^2-4a+9} = \dfrac{a(a-1)^2}{18(a^2-4a+9)} \geq 0,$$
$$\therefore \dfrac{1}{a^2-4a+9} \leq \dfrac{a}{18} + \dfrac{1}{9}.$$
So$:$ $$ \sum \dfrac{1}{a^2-4a+9} \leq \dfrac{a+b+c}{18} + \dfrac{1}{3} = \dfrac{7}{18}.$$
Equality occur when $a:b:c=1:0:0$ or any permution.
Update. Let
$$f(a) = (\,ma + n\,)(\,a^2-4a+9\,) - 1.$$
We will try to find $m,\,n$ such that $f(a) \geqslant 0$ for all $a \in [\,0,1\,].$
Proof 1. Let $a = 0$ and $a =1$ we get
$$\{9n = 1,\; 6(m+n) = 1\}.$$
Solve equation we get $m = \dfrac 1 {18},$ $n = \dfrac 1 9.$
Proof 2. Write inequality as
$$f(a) = a \Big[\,a^2m-(4m-n)a+9m-4n\,\Big] +9n-1.$$
It's easy choose $n = \dfrac{1}{9},$ we get
$$f(a) = a \Big[\,ma^2-(4m-n)a+9m-4n\,\Big].$$
For a quadratic polynomial $ma^2-(4m-n)a+9m-4n,$ we have$:$
$$\Delta  = -(10m+n)(2m-n) = -\left(10m+\dfrac 1 9\right)\left(2m-\dfrac 1 9\right).$$
We need $m\geqslant 0.$
So choose $2m = \dfrac{1}{9}$ which means $m = \dfrac{1}{18}.$
