Finding the maximum value of $\sum_{cyc}\frac{1}{a^2-4a+9}$ s.t. $a+b+c =1$

Question

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.

Answer 1

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 $m = \dfrac{1}{18}.$

Answer 2

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.