\(29xyz=\left(x+y+z\right)^3+x^2+y^2+z^2+4\ge27xyz+3\sqrt[3]{\left(xyz\right)^2}+4\)
\(\Leftrightarrow2xyz-3\sqrt[3]{\left(xyz\right)^2}-4\ge0\)
Đặt \(\sqrt[3]{xyz}=t>0\Rightarrow2t^3-3t^2+4\ge0\)
\(\Leftrightarrow\left(t-2\right)\left(2t^2+t+2\right)\ge0\)
\(\Leftrightarrow t\ge2\Leftrightarrow xyz\ge8\)
\(\Rightarrow xyz_{min}=8\) khi \(x=y=z=2\)