Inequality with $abc=1$ Let $a;b;c$ be positive real numbers and $abc=1$
Find maximum value of: $P=\dfrac{1}{\sqrt{a^2+2b^2}}+\dfrac{1}{\sqrt{b^2+2c^2}}+\dfrac{1}{\sqrt{c^2+2a^2}}$
I tried to use Cauchy-Schwarz: $a^2+2b^2 \geq \dfrac{1}{3}(a+2b)^2$
Then: $P \leq \dfrac{\sqrt{3}}{a+2b}+\dfrac{\sqrt{3}}{b+2c}+\dfrac{\sqrt{3}}{c+2a}$
So, what should I do next? Thank you.

There is no maximum .
Note that when $a\to 0^+,b\to 0^+,c\to \infty$ we get  $P\to \infty$ .This is because the $\sqrt{a^2+2b^2}\to 0$
Note that we chose $c\to \infty$ because $c=\frac{1}{ab}$

Now even if you are looking for the minimum there is no minimum .
Note  again that  $a\to 0^+,b\to \infty,c\to \infty$ we see that $P\to 0$.
