$A=\dfrac{a^3}{b+2c}+\dfrac{b^3}{c+2a}+\dfrac{c^3}{a+2b}$
$\Leftrightarrow A=\dfrac{a^4}{ab+2ac}+\dfrac{b^4}{bc+2ab}+\dfrac{c^4}{ac+2bc}$
Áp dụng bất đẳng thức Cauchy Schwarz, ta có
$A \ge \dfrac{(a^2+b^2+c^2)^2}{3(ab+ac+bc)}$
Mà $\dfrac{(a^2+b^2+c^2)^2}{3(ab+ac+bc)} \ge \dfrac{1}{9}(a+b+c)^2$. Thật vậy
$a^2+b^2+c^2 \ge ab+ac+bc$
$3(a^2+b^2+c^2) \ge (a+b+c)^2$
$\Rightarrow 3(a^2+b^2+c^2)^2 \ge (ab+ac+bc)(a+b+c)^2$
$\Leftrightarrow \dfrac{(a^2+b^2+c^2)^2}{3(ab+ac+bc)} \ge \dfrac{1}{9}(a+b+c)^2$
Vậy, $\dfrac{a^3}{b+2c}+\dfrac{b^3}{c+2a}+\dfrac{c^3}{a+2b} \ge \dfrac{1}{9}(a+b+c)^2$