Let $a,b,c>0$ and $a+b+c=3$. Prove that $$\frac{1}{1+2b^2c}+\frac{1}{1+2c^2a}+\frac{1}{1+2a^2b}\ge1.$$
My attempts:
1) I use Titu's Lemma:
$$\frac{1}{1+2b^2c}+\frac{1}{1+2c^2a}+\frac{1}{1+2a^2b}\ge\frac{(1+1+1)^2}{3+2(a^2b+b^2c+c^2a)}$$
2) I use AM-GM:
$$\frac{1}{1+2b^2c}=1-\frac{2b^2c}{1+2b^2c}\ge1-\frac{bc}{\sqrt{2c}}$$
