\(A=x\left(x-1\right)+y\left(y-1\right)=x^2+y^2-\left(x+y\right)\)
\(A\ge\frac{1}{2}\left(x+y\right)^2-\left(x+y\right)\)
Đặt \(x+y=a\Rightarrow a\ge6\)
\(A\ge\frac{1}{2}a^2-a=\frac{1}{2}a^2-a-12+12\)
\(A\ge\frac{1}{2}\left(a-6\right)\left(a+4\right)+12\ge12\)
\(A_{min}=12\) khi \(a=6\) hay \(x=y=3\)