summation - Inequality with a condition

If $a,b,c$ positives and $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} =3,$ I have to prove $\frac {1}{2 a^2+b^2} +\frac {1}{2 b^2+c^2} +\frac {1}{2 c^2+a^2} \le 1.$
Since we have $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} =3$, then implying Titu's lemma, we have:

$ \frac{2}{a^2} +\frac{1}{b^2} =\frac{1}{a^2} +\frac{1}{a^2} +\frac{1}{b^2} \ge \frac {9}{2 a^2+b^2},$
so:
$ \frac {1}{2 a^2+b^2} +\frac {1}{2 b^2+c^2} +\frac {1}{2 c^2+a^2} \le \frac{1}{3} ( \frac {1}{a^2} +\frac {1}{b^2} +\frac {1}{c^2}).$

It is enough to prove:
$\frac {1}{a^2} +\frac {1}{b^2} +\frac {1}{c^2} \le 3= \frac{1}{3} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
and I stuck. Thank you

Answer

By AM-GM $$1-\sum_{cyc}\frac{1}{2a^2+b^2}=\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-\sum_{cyc}\frac{1}{2a^2+b^2}=$$
$$=\frac{1}{9}\sum_{cyc}\left(\frac{1}{a^2}+\frac{2}{ab}-\frac{9}{2a^2+b^2}\right)=\sum_{cyc}\frac{4a^3+2ab^2+b^3-7a^2b}{9a^2b(2a^2+b^2)}\geq$$
$$\geq\sum_{cyc}\frac{7\sqrt[7]{(a^3)^4(ab^2)^2b^3}-7a^2b}{9a^2b(2a^2+b^2)}=0.$$
Also, we can use C-S and AM-GM:
$$\frac{1}{a^2}+\frac{2}{ab}=\frac{1}{a^2}+\frac{4}{2ab}\geq\frac{(1+2)^2}{a^2+2ab}\geq\frac{9}{a^2+2\cdot\frac{a^2+b^2}{2}}=\frac{9}{2a^2+b^2}.$$