Question

[Lizy]

Tìm min `A=x^2 + xy + y^2- 3(x+y) +3`.

Answer 1

[Toru]

\(A=x^2+xy+y^2-3(x+y)+3\\2A=2x^2+2xy+2y^2-6(x+y)+6\\=(x^2+2xy+y^2)-4(x+y)+4+(x^2-2x+1)+(y^2-2y+1)\\=(x+y)^2-4(x+y)+4+(x-1)^2+(y-1)^2\\=(x+y-2)^2+(x-1)^2+(y-1)^2\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x+y-2\right)^2\ge0\forall x,y\\\left(x-1\right)^2\ge0\forall x\\\left(y-1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x+y-2\right)^2+\left(x-1\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

\(\Rightarrow2A\ge0\forall x,y\)

\(\Rightarrow A\ge0\forall x,y\)

Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}x+y-2=0\\x-1=0\\y-1=0\end{matrix}\right.\Rightarrow x=y=1\)

Vậy \(Min_A=0\) khi \(x=y=1\).

\(\text{#}Toru\)

Answer 2

[Nguyễn Việt Lâm]

\(2A=2x^2+2y^2+2xy-6x-6y+6\)

\(2A=\left(x+y\right)^2-4\left(x+y\right)+4+\left(x-1\right)^2+\left(y-1\right)^2\)

\(2A=\left(x+y-2\right)^2+\left(x-1\right)^2+\left(y-1\right)^2\)

Do \(\left\{{}\begin{matrix}\left(x+y-2\right)^2\ge0\\\left(x-1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\) ;\(\forall x;y\)

\(\Rightarrow2A\ge0\Rightarrow A\ge0\)

Vậy \(A_{min}=0\) khi \(\left\{{}\begin{matrix}x+y-2=0\\x-1=0\\y-1=0\end{matrix}\right.\) hay \(\left(x;y\right)=\left(1;1\right)\)