Maximize $\frac{1}{\sqrt{2a^2+b^2}}+\frac{1}{\sqrt{2a^2+c^2}}+\frac{1}{\sqrt{2c^2+a^2}}$ under some condition on $(a,b,c)$

Question

For $a,b,c$ are postive real number satisfy $7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)+2015$. Find Maximize $$P=\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}$$

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I find 2 Max is: $Max=\sqrt{\frac{2015}{3}}$ at $a=b=c=\sqrt {\frac{3}{2015}}$ and $Max=\sqrt{\frac{6045}{3}} $ at $a=b=c=\sqrt{\frac{3}{6045}}$ Which right ?

Answer 1

By C-S $$\sum_{cyc}\frac{1}{\sqrt{3(2a^2+b^2)}}=\sum_{cyc}\frac{1}{\sqrt{(2+1)(2a^2+b^2)}}\leq\sum_{cyc}\frac{1}{2a+b}\leq\frac{1}{9}\sum_{cyc}\left(\frac{2^2}{2a}+\frac{1^2}{b}\right)=$$ $$=\frac{1}{3}\sum_{cyc}\frac{1}{a}\leq\sqrt{\frac{1}{3}\sum_{cyc}\left(\frac{7}{a^2}-\frac{6}{ab}\right)}=\sqrt{\frac{2015}{3}}.$$ The equality occurs for $\sum\limits_{cyc}\left(\frac{7}{a^2}-\frac{6}{ab}\right)=2015$ and $a=b=c$.

Thus, the equality occurs.

Id est, the answer is $\sqrt{\frac{2015}{3}}$.