\(0\le x;y;z\le1\Rightarrow\left\{{}\begin{matrix}y^4\le y^2\\z^{2020}\le z^2\\\end{matrix}\right.\)
\(\Rightarrow G\le x^2+y^2+z^2-xy-yz-zx\)
Mặt khác do \(x;y;z\in\left[0;1\right]\)
\(\Rightarrow xyz+\left(1-x\right)\left(1-y\right)\left(1-z\right)\ge0\)
\(\Rightarrow xy+yz+zx-\left(x+y+z\right)+1\ge0\)
\(\Rightarrow-\left(xy+yz+zx\right)\le1-\left(x+y+z\right)\)
\(\Rightarrow G\le x^2+y^2+z^2-\left(x+y+z\right)+1\)
\(\Rightarrow G\le x\left(x-1\right)+y\left(y-1\right)+z\left(z-1\right)+1\)
Do \(x;y;z\in\left[0;1\right]\Rightarrow\left\{{}\begin{matrix}x\left(x-1\right)\le0\\y\left(y-1\right)\le0\\z\left(z-1\right)\le0\end{matrix}\right.\)
\(\Rightarrow G\le1\)
\(G_{max}=1\) khi \(\left(x;y;z\right)=\left(0;1;1\right);\left(0;0;1\right)\) và các hoán vị