$\frac{a^2}{a+2b^2}=a-\frac{2ab^2}{a+2b^2}\geq a-\frac{2\sqrt[3]{a^2b^2}}{3}$
$\Rightarrow \sum \frac{a^2}{a+2b^2}\geq a+b+c-\frac{2}{3}(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2})$
$Holder$ $\frac{(a+b+c)^4}{3}\geq 3(ab+bc+ca)^2\geq (\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2})^3$
$\Rightarrow \sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2}\leq 3$
$\Rightarrow \sum \frac{a^2}{a+2b^2}\geq 1$.