a.
\(xy\le\frac{\left(x+y\right)^2}{4}=\frac{2^2}{4}=1\)
Vì x,y>0 nên \(xy>0\)
Vậy \(0< xy\le1\)
b.
\(A=x^2y^2\left(x^2+y^2\right)=\frac{xy}{2}\cdot2xy\left(x^2+y^2\right)\le\frac{\left(x+y\right)^2}{8}\cdot\frac{\left(x+y\right)^4}{4}=\frac{2^6}{8\cdot4}=2\)
Vậy \(A_{max}=2\Leftrightarrow x=y=1\)