Ta có: \(\left(x+y-\frac{1}{2}\right)^2\ge0\forall x,y\)
\(\left|x-y+\frac{1}{3}\right|\ge0\forall x,y\)
\(\Rightarrow\left(x+y-\frac{1}{2}\right)^2+\left|x-y+\frac{1}{3}\right|\ge0\)
Dấu " = " xảy ra khi:
\(\left(x+y-\frac{1}{2}\right)^2=\left|x-y+\frac{1}{3}\right|=0\)
\(\Rightarrow x+y-\frac{1}{2}=x-y+\frac{1}{3}=0\)
\(\Rightarrow x=\frac{5}{12};y=\frac{1}{12}\)
\(\Rightarrow\) Không tồn tại \(x,y\) nguyên thỏa mãn đề.