\(H=x^2+2y^2-2xy+6y+2023\\=(x^2-2xy+y^2)+(y^2+6y+9)+2014\\=(x-y)^2+(y^2+2\cdot y\cdot3+3^2)+2014\\=(x-y)^2+(y+3)^2+2014\)
Ta thấy: \(\left(x-y\right)^2\ge0\forall x;y\)
\(\left(y+3\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-y\right)^2+\left(y+3\right)^2\ge0\forall x;y\)
\(\Rightarrow H=\left(x-y\right)^2+\left(y+3\right)^2+2014\ge2014\forall x;y\)
Dấu \("="\) xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=-3\end{matrix}\right.\)
\(\Leftrightarrow x=y=-3\)
Vậy \(Min_H=2014\) khi \(x=y=-3\)
\(H=x^2+2y^2-2xy+6y+2023\)
\(2H=2x^2+4y^2-4xy+12y+4046\)
\(2H=4y^2-4y\left(x-3\right)+\left(x-3\right)^2-\left(x-3\right)^2+2x^2+4046\)
\(2H=\left(2y-x+3\right)^2+x^2+6x+9+4028\)
\(H=\dfrac{1}{2}\left[\left(2y-x+3\right)^2+\left(x+3\right)^2\right]+2014\)
Vì \(\left(2y-x+3\right)^2+\left(x+3\right)^2\ge0\forall x,y\)
\(MinH=2014\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-3\end{matrix}\right.\)
