Đáp án: `min G=-5` khi `(x;y;z)=(1;2;3)`
Giải thích các bước giải:
`G=2x^2+2y^2+z^2+2xy-2xz-2yz-2x-4y`
`=x^2+y^2+z^2+2xy-2xz-2yz+x^2-2x+1+y^2-4y+4-5`
`=[(x^2+2xy+y^2)-2.z.(x+y)+z^2]+(x^2-2x+1)+(y^2-4y+4)-5`
`=[(x+y)^2-2.z.(x+y)+z^2]+(x-1)^2+(y-2)^2-5`
`=(x+y-z)^2+(x-1)^2+(y-2)^2-5`
Vì `(x+y-z)^2≥0∀x;y;z\text( )(x-1)^2≥0∀x\text( )(y-2)^2≥0∀y`
`->(x+y-z)^2+(x-1)^2+(y-2)^2≥0∀x;y;z`
`->(x+y-z)^2+(x-1)^2+(y-2)^2-5≥-5∀x;y;z`
Dấu `'='` xảy ra `<=>`$\begin{cases}x+y-z=0\\x-1=0\\y-2=0\end{cases}$`<=>`$\begin{cases}1+2=z\\x=1\\y=2\end{cases}$
`<=>`$\begin{cases}z=3\\x=1\\y=2\end{cases}$
Vậy `min G=-5` khi `(x;y;z)=(1;2;3)`
