đặt a=x^{3},b=y^{3},c=z^{3} khi đó xyz=1P= \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\leq1dấu "=" \Leftrightarrow a=b=c=1
đặt
a=x^{3},b=y^{3},c=z^{3} khi đó
xyz=1P=
\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\leq1dấu "="
\Leftrightarrow a=b=c=1