Giải thích các bước giải:
$a+b+c\ge 3abc\to \dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge 3$
Đặt $\dfrac 1a=x, \dfrac 1b=y,\dfrac 1c=z\to x+y+z\ge 3$
$\to P=\dfrac{\dfrac{1}{a^3}}{\dfrac{1}{c}+\dfrac{2}{b}}+\dfrac{\dfrac{1}{b^3}}{\dfrac{1}{a}+\dfrac{2}{c}}+\dfrac{\dfrac{1}{c^3}}{\dfrac{1}{b}+\dfrac{2}{a}}$
$\to P=\dfrac{x^3}{z+2y}+\dfrac{y^3}{x+2z}+\dfrac{z^3}{y+2x}$
$\to P((x+2y)+(x+2y)+(x+2y))(1+1+1)=(\dfrac{x^3}{z+2y}+\dfrac{y^3}{x+2z}+\dfrac{z^3}{y+2x})((x+2y)+(x+2y)+(x+2y))(1+1+1)$
$\to 3P((x+2y)+(x+2y)+(x+2y))\ge (\sqrt[3]{\dfrac{x^3}{z+2y}.(z+2y).1}+\sqrt[3]{\dfrac{y^3}{x+2z}.(x+2z).1}+\sqrt[3]{\dfrac{z^3}{y+2x}.(y+2x).1})^3$
$\to 3P(3(x+y+z))\ge (x+y+z)^3$
$\to P\ge \dfrac{(x+y+z)^2}{9}\ge \dfrac{xy+yz+zx}{9}\ge \dfrac{3}{9}=\dfrac 13$
