Ta có:$8=(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc\ge\dfrac{8}{9}(a+b+c)(ab+bc+ca)$
$\Rightarrow (a+b+c)(ab+bc+ca)\le 9$.
Ta có:
$P=\dfrac{1}{\sqrt[3]{abc}}+\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}$
$\ge\dfrac{1}{\sqrt[3]{abc}}+\dfrac{3}{a+b+c}$
$\ge2\sqrt{\dfrac{1}{\sqrt[3]{abc}}.\dfrac{3}{a+b+c}}$
$=\dfrac{2\sqrt3}{\sqrt[6]{abc(a+b+c)^3}}$
$\ge\dfrac{2\sqrt3}{\sqrt[6]{\dfrac{(ab+bc+ca)^2(a+b+c)^2}{3}}}\ge2$.
Vậy $\min P=2 \Leftrightarrow a=b=c=1$.