đặt $a=x^{3},b=y^{3},c=z^{3}$ khi đó $xyz=1$P=$ \sum \frac{1}{x^{3}+y^{3}+1}$
ta có $x^{3}+y^{3}+1\geq xy(x+y)+xyz=xy(x+y+z)$
$\Rightarrow \frac{1}{x^{3}+y^{3}+1}\leq \frac{1}{xy(x+y+z)}=\frac{z}{x+y+z}$
$\Rightarrow P\leq1$
dấu "=" $\Leftrightarrow a=b=c=1$