For classical inequalities lovers

First we show that $3$ is an upper bound;

Apply AM-GM:

$\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}=\frac{1}{a+a+b}+\frac{1}{b+b+c}+\frac{1}{c+c+a}$

$\le \left(\frac{1}{3(a^2b)^{1/3}}+\frac{1}{3(b^2c)^{1/3}}+\frac{1}{3(c^2a)^{1/3}}\right)$

By Chebyshev:

$\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}{3}\le \left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)$

Hence,

$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le3\sqrt{3}$

Using Holder one obtains:

$\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)(1+1+1)\ge\left(\underbrace{\frac{1}{(a^2b)^{1/3}}+\frac{1}{(b^2c)^{1/3}}+\frac{1}{(c^2a)^{1/3}}}_{=S}\right)^3$

Make use of $(2)$ :

$9\cdot3\sqrt{3}\cdot3\ge S^3$

Or,

$(3^9)^{1/2}\ge S^3$

Equivalently,

$S\le\sqrt{27}$

Therefore the square of our sum in $(1)$ is bounded above by:

$\frac{S^2}{9}=3$

And setting $a^2=b^2=c^2=\frac{1}{3}$ this bound can be attained