\(\left(x+y\right)^2\Rightarrow4xy\Rightarrow\left(x+y\right)^3+\left(x+y\right)^2\ge\left(x+y\right)^3+4xy\ge2\)
\(\Rightarrow\left(x+y\right)^3+\left(x+y\right)^2-2\ge0\)
\(\Rightarrow\left(x+y-2\right)\left[\left(x+y+1\right)^2+1\right]\ge0\)
\(\Rightarrow x+y\ge2\) \(\Rightarrow x^2+y^2\ge\frac{1}{2}\left(x+y\right)^2\ge2\)
Ta có: \(A=3\left(x^2+y\right)^2-3x^2y^2-2\left(x^2+y^2\right)+1\)
\(A\ge3\left(x^2+y^2\right)^2-\frac{3}{4}\left(x^2+y^2\right)^2-2\left(x^2+y^2\right)+1=\frac{9}{4}\left(x^2+y^2\right)^2-2\left(x^2+y^2\right)+1\)
\(A\ge\frac{9}{4}\left(x^2+y^2-2\right)\left(x^2+y^2+\frac{10}{9}\right)+6\ge6\)
\(A_{min}=6\) khi \(x=y=1\)
