\(\left|x+y\right|+\left(y-4\right)^2+2013\left(1\right)\)
Vì \(\left|x+y\right|\ge0\forall x,y;\left(y-4\right)^2\ge0\forall y\)
\(\Rightarrow\left(1\right)\ge2013\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=4\end{matrix}\right.\)