\(A=x^2+xy+y^2-3x-3y+3002\)
\(=\left(x^2-2x+1\right)+\left(y^2-2x+1\right)+\left(xy-x-y+1\right)+2009\)
\(=\left(x-1\right)^2+\left(y-1\right)^2+\left(x-1\right)\left(y-1\right)+2009\)
\(=\left(x-1\right)^2+\dfrac{1}{4}\left(y-1\right)^2+2.\left(x-1\right).\dfrac{1}{2}\left(y-1\right)+\dfrac{3}{4}\left(y-1\right)^2+2009\)
\(=\left[\left(x-1\right)+\dfrac{1}{2}\left(y-1\right)\right]^2+\dfrac{3}{4}\left(y-1\right)^2+2009\)
Ta thấy : \(\left\{{}\begin{matrix}\left[\left(x-1\right)+\dfrac{1}{2}\left(y-1\right)\right]^2\ge0\forall x;y\\\dfrac{3}{4}\left(y-1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow A=\left[\left(x-1\right)+\dfrac{1}{2}\left(y-1\right)\right]^2+\dfrac{3}{4}\left(y-1\right)^2+2009\ge2009\)
Dấu "=" xảy ra <=> x = y = 1
Vậy x = y = 1 thì A đạt GTNN là 2009
