Đặt \(A=x^2+y^2+xy+3x+3y+2018\)
\(4.A=4x^2+4y^2+4xy+12x+12y+8072\)
\(4.A=\left(4x^2+4xy+y^2\right)+3y^2+12x+12y+8072\)
\(4.A=\left[\left(2x+y\right)^2+2\left(2x+y\right).3+9\right]+3\left(y^2+2y+1\right)+8060\)
\(4.A=\left(2x+y+3\right)^2+3\left(y+1\right)^2+8060\)
Mà \(\left(2x+y+3\right)^2\ge0\forall x;y\)
\(\left(y+1\right)^2\ge0\forall y\)\(\Rightarrow3\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow4.A\ge8060\)
\(\Leftrightarrow A\ge2015\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}2x+y+3=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases}}\)
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