Giải thích các bước giải:
Đặt $a=x^3, b=y^3, c=z^3\rightarrow xyz=1$
$\rightarrow a+b+1=x^3+y^3+xyz\ge xy(x+y)+xyz=xy(x+y+z)$
Tương tự :
$\rightarrow \begin{cases}b+c+1=y^3+z^3+xyz\ge yz(y+z+x)\\c+a+1=z^3+x^3+xyz\ge zx(x+z+y)\end{cases}$
$\rightarrow P=\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{c+a+1}=\dfrac{1}{x^3+y^3+xyz}+\dfrac{1}{y^3+z^3+xyz}+\dfrac{1}{z^3+x^3+zyx}$
$\rightarrow P\le \dfrac{1}{xy(x+y+z)}+\dfrac{1}{yz(y+z+x)}+\dfrac{1}{zx(z+x+y)}$
$\rightarrow P\le \dfrac{z}{x+y+z}+\dfrac{x}{y+z+x}+\dfrac{y}{z+x+y}$
$\rightarrow P\le 1\rightarrow đpcm$
