$A=\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}$
Ta có: $\frac{a^2}{a+2b^2}+(a+2b^2)\geq 2a$ (BĐT Cô-si)
$\frac{b^2}{b+2c^2}+(b+2c^2)\ge 2b$
$\frac{c^2}{c+2a^2}+(c+2a^2)\ge2c$
$\Rightarrow A+a+b+c+2(a^2+b^2+c^2)\ge2(a+b+c)$
$\Rightarrow A\ge 3-2(a^2+b^2+c^2)$
Theo BĐT Bunhia: $(a+b+c)^2\leq3(a^2+b^2+c^2)$
$\Rightarrow a^2+b^2+c^2\ge1$
$\Rightarrow A \ge 1 (đpcm)$