Xét \(A=x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)^3\left(x^2+y^2+z^2+2xy-xz-yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
\(\Rightarrow2A=2\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
\(=\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2xz-2yz\right)\)
\(=\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]\)
Vì \(x+y+z\ge0\) ; \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\) với mọi \(x,y,z\)
\(\Rightarrow2A\ge0\)
\(\Rightarrow A\ge0\)
\(\Rightarrow x^3+y^3+z^3\ge3xyz\)
Vậy nếu \(x+y+z\ge0\) thì \(x^3+y^3+z^3\ge3xyz\)