$tu gia thiet va bdt A= \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\geq \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} ta co$
$ 28A-2013\leq 4A<=> A\leq \frac{2013}{24}$
$ap dung bdt BCS co P\leq 3(\frac{1}{5a^{2}+2ab+b^{2}}+\frac{1}{5b^{2}+2bc+c^{2}}+\frac{1}{5c^{2}+2ac+a^{2}})$
$\frac{1}{5a^{2}+2ab+b^{2}}\leq \frac{1}{4}(\frac{1}{4a^{2}}+\frac{1}{(a+b)^{2}})\leq \frac{1}{16a^{2}}+\frac{1}{16ab}$
$ tu do tuong tu vs 2 hang tu kia dk $
$P\leq \frac{1}{16}(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})=\frac{A}{16}+\frac{28A-2013}{4.16}=\frac{A}{2}+\frac{2013}{64}\leq \frac{4697}{64}$
$vay Pmax= \frac{4697}{64} dat dk khi a=b=c=\frac{\sqrt{2013} }{6\sqrt{2} }$