Đặt vế trái là P
Ta có: \(P\le x^2y+y^2z+z^2x+xyz\)
Không mất tính tổng quát, giả sử \(x=mid\left\{x;y;z\right\}\Rightarrow\left(x-y\right)\left(x-z\right)\le0\)
\(\Leftrightarrow x^2+yz\le xy+xz\)
\(\Rightarrow x^2y+y^2z\le xy^2+xyz\)
\(\Rightarrow P\le xy^2+z^2x+2xyz=x\left(y^2+z^2+2yz\right)=x\left(y+z\right)^2\)
\(\Rightarrow P\le\frac{1}{2}.2x\left(y+z\right)\left(y+z\right)\le\frac{1}{2}\left(\frac{2x+y+z+y+z}{3}\right)^3=\frac{4}{27}\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\frac{1}{3};0;\frac{2}{3}\right)\)
