Find $\min\big\{ \lfloor xy + \frac{1}{xy} \rfloor \,\Big|\, (x+1)(y+1)=2 ,\, 0I am invited to calculate the minimum of the following set:
$\big\{ \lfloor xy + \frac{1}{xy} \rfloor \,\Big|\, (x+1)(y+1)=2 ,\, 0<x,y \in \mathbb{R} \big\}$.
Is there any idea?
(The question changed because there is no maximum for the set (as proved in the following answers) and I assume that the source makes mistake)

There is a minimum of $6$ achieved at $x=y=\sqrt{2}-1$
to prove this, start with the identity
$\left(x-\frac{1}{x}\right)\left(y-\frac{1}{y}\right)+\left(\frac{y}{x}+\frac{x}{y}\right)=\left(x y+\frac{1}{x y}\right)$
a simple verification shows that $\left(x-\frac{1}{x}\right)\left(y-\frac{1}{y}\right)=4$
on the other hand $\frac{y}{x}+\frac{x}{y}\geq 2$
it follows that $x y+\frac{1}{x y}\geq 6$
