Ta có $a^2-ab+b^2\geq \frac{1}{3}(a^2+ab+b^2)$
Đặt $B=\frac{(a+b)(a^2-ab+b^2)^2}{a^2+ab+b^2}+\frac{(b+c)(b^2-bc+c^2)^2}{b^2+bc+c^2}+\frac{(c+a)(c^2-ca+a^2)^2}{c^2+ca+a^2}$
$\Rightarrow B=\sum \frac{(a+b)(a^2-ab+b^2)^2}{a^2+ab+b^2} \geq \sum \frac{\frac{1}{3}(a+b)(a^2-ab+b^2)(a^2+ab+b^2)}{a^2+ab+b^2}=\frac{1}{3}(\sum (a+b)(a^2-ab+b^2))=\frac{1}{3}(\sum a^3+b^3)=\frac{2}{3}(a^3+b^3+c^3)$
Mà do $\frac{abc}{ab+bc+ca}\geq \frac{1}{9} \Rightarrow 9abc\geq ab+bc+ca \Rightarrow ab+bc+ca \leq 9\sqrt{(\frac{ab+bc+ca}{3})^3} \Rightarrow ab+bc+ca \geq \frac{1}{3}$
áp dụng BĐT AM-GM ta có $a^3+b^3+c^3 \geq 3abc\geq \frac{abc}{ab+bc+ca}$
$\Rightarrow B\geq \frac{2}{3}\frac{abc}{ab+bc+ca}$
$\Rightarrow A\geq \frac{2}{3}\frac{abc}{ab+bc+ca}+\frac{2(ab+bc+ca)}{243abc}=\frac{2}{243}(\frac{81abc}{ab+bc+ca}+\frac{(ab+bc+ca)}{abc})\geq \frac{2}{243}.2\sqrt{\frac{81abc}{ab+bc+ca}.\frac{(ab+bc+ca)}{abc}}=\frac{4}{27}$
Dấu bằng xảy ra khi $a=b=c=\frac{1}{3}$