Áp dụng bđt \(\frac{a^2}{m}+\frac{b^2}{n}+\frac{c^2}{p}\ge\frac{\left(a+b+c\right)^2}{m+n+p}\) được
\(G\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}=\frac{2}{2}=1\)
\(G\ge1\Rightarrow MinG=1\Leftrightarrow\hept{\begin{cases}x=y=z>0\\x+y+z=2\end{cases}\Leftrightarrow}x=y=z=\frac{2}{3}\)