\(A=x^3+y^3+x^2+y^2\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+\left(x+y\right)^2-2xy\)
\(=1^3-3xy+1^2-2xy=2-5xy\)
Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1^2}{4}=\frac14\)
=>\(-5xy\ge-\frac54\forall x,y\)
=>\(A=-5xy+2\ge-\frac54+2=\frac34\forall x,y\)
Dấu '=' xảy ra khi \(x=y=\frac12\)