Từ giả thiết ta có:$28(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})=4(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca})+2013\le4(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})+2013$
$\Rightarrow \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\le\dfrac{671}{8}$
Ta có:
$\dfrac{1}{\sqrt{5a^2+2ab+b^2}}=\dfrac{8\sqrt3}{\sqrt{671}}\sqrt{\dfrac{1}{5a^2+2ab+b^2}.\dfrac{671}{192}}$
$\le\dfrac{4\sqrt3}{\sqrt{671}}\left(\dfrac{1}{5a^2+2ab+b^2}+\dfrac{671}{192}\right)$
$\le\dfrac{4\sqrt3}{\sqrt{671}}\left[\dfrac{1}{64}\left(\dfrac{5}{a^2}+\dfrac{2}{ab}+\dfrac{1}{b^2}\right)+\dfrac{671}{192}\right]$
$=\dfrac{\sqrt3}{16\sqrt{671}}\left(\dfrac{5}{a^2}+\dfrac{2}{ab}+\dfrac{1}{b^2}\right)+\dfrac{\sqrt{671}}{16\sqrt3}$
Tương tự: $\dfrac{1}{\sqrt{5b^2+2bc+c^2}}\le\dfrac{\sqrt3}{16\sqrt{671}}\left(\dfrac{5}{b^2}+\dfrac{2}{bc}+\dfrac{1}{c^2}\right)+\dfrac{\sqrt{671}}{16\sqrt3}$
$\dfrac{1}{\sqrt{5c^2+2ca+a^2}}\le\dfrac{\sqrt3}{16\sqrt{671}}\left(\dfrac{5}{c^2}+\dfrac{2}{ca}+\dfrac{1}{a^2}\right)+\dfrac{\sqrt{671}}{16\sqrt3}$
Cộng các BĐT trên ta được:
$P\le\dfrac{6\sqrt3}{16\sqrt{671}}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+\dfrac{2\sqrt3}{16\sqrt{671}}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)+\dfrac{3\sqrt{671}}{16\sqrt3}$
$\le\dfrac{\sqrt3}{2\sqrt{671}}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+\dfrac{\sqrt{2013}}{16}\le\dfrac{\sqrt{2013}}{8}$
Vậy $\max P=\dfrac{\sqrt{2013}}{8} \Leftrightarrow a=b=c=\dfrac{2\sqrt6}{\sqrt{671}}$