$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{3}\geq \frac{2\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^2}{3}\geq \frac{\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^2+3\left ( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \right )}{3} \Rightarrow 1\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
$\sum_{}^{} \frac{}{}\frac{1}{\sqrt{3(a^2+a^2+b^2)}}\leq \sum_{}^{}\frac{1}{a+a+b}\leq \sum_{}^{}\frac{1}{9}\left ( \frac{2}{a}+\frac{1}{b} \right ) =\frac{1}{3}\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right ) \leq \frac{1}{3} \Leftrightarrow a=b=c=3$