Vì $a,b,c>0$ nên áp dung bddoooooooooooooooooooooo
$\frac{a^2}{a+2b}+\frac{a+2b}{9}\geq\frac{2}{3}a $
$\Leftrightarrow \frac{a^2}{a+2b}\geq \frac{5a-2b}{9}(1)$
Tương tự ta co:
$\frac{b^2}{b+2c}\geq \frac{5b-2c}{9}(2)$
$\frac{c^2}{c+2a}\geq \frac{5c-2a}{9}(3)$
Từ $(1),(2),(3)$ ta co:$\frac{a^2}{a+2b}+\frac{b^2}{b+2c}+\frac{c^2}{c+2a}\geq \frac{3(a+b+c)}{9}=1 (vì a+b+c=3)$
Dấu bằn xảy ra $\Leftrightarrow a=b=c=1$