The maximum value problem of basic inequality Let x > 0, Y > 0, and XY (x + y + 1) = 1, then the minimum value of (x + y) (y + 1) is

The maximum value problem of basic inequality Let x > 0, Y > 0, and XY (x + y + 1) = 1, then the minimum value of (x + y) (y + 1) is

The minimum value is 2
∵xy(x+y+1)=1,∴y(x+y+1)=1/x.
therefore:
(x+y)(y+1)=xy+x+y^2+y=x+y(x+y+1)=x+1/x≧2.