Prove that $\sum_{cyc} \frac{1}{5a^2-4a+11} \le \frac{1}{4}$

Question

The problem

Given that $a+b+c=3, $ Prove that \begin{align} \sum_{cyc} \frac{1}{5a^2-4a+11} \le \frac14 \end{align}

Attempts

So one apparent notice is that equality is achieved at $a=b=c=1$. Now I intend to calculate the derivative of the fraction,

so

\begin{align} (\frac{1}{5a^2-4a+11})' = \frac{-10a+4} {(5a^2-4a+11)^2} \end{align}

which means each individual fraction achieve the highest value at $a=0,4$ instead of $1$.

In one last hope, I tried to convert $c=3-a-b$ but obviously since there's no equality at any $a=b$ (other than $1$), the scheme doesn't work.

So is there any other method I can try?

Any help is appreciated.