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Vì $a,b,c>0$ nên áp dung bddt cauchy cho 2 số ta co:$\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 $\Leftrightarrow a=b=c=1$

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 $\Leftrightarrow a=b=c=1$