Biểu thức xyz chỉ có max, ko có min
\(1\ge2xyz+xy+yz+zx\ge2xyz+3\sqrt[3]{\left(xyz\right)^2}\)
Đặt \(\sqrt[3]{xyz}=t>0\Rightarrow2t^3+3t^2-1\le0\)
\(\Leftrightarrow\left(t+1\right)^2\left(2t-1\right)\le0\)
\(\Leftrightarrow2t-1\le0\Rightarrow t\le\frac{1}{2}\)
\(\Rightarrow\sqrt[3]{xyz}\le\frac{1}{2}\Rightarrow xyz\le\frac{1}{8}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)