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)
 A: 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 $x y+\frac{1}{x y}\geq 6$
A: This needs further verification. 
I believe that the maximum does not exist since the set is not bounded above. 
Suppose $x, y \gt 0$ satisfies $ (x + 1)(y + 1) = 2 $. Then we can conclude that 


*

*$ xy + x+ y = 1 $

*$ 1 + \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{xy}  $ (dividing by $xy$)


Subtracting equation $2$ from $1$ we can rearrange this to $$ xy + \dfrac{1}{xy} = 2 + \left({\dfrac{1}{x} - x}\right  ) +  \left({\dfrac{1}{y} - y}\right  ) $$  
Now notice that the expression $\left({\dfrac{1}{t} - t}\right  )$ can be made arbitrarily large by making $t$ arbitrarily small. Hence the floor function  $ \lfloor xy + \frac{1}{xy} \rfloor  $ is also not bounded above and so the maximum of the given set cannot exist. 
A: Solve for one variable using the equation $(x+1)(y+1)=2$.  Solving for $x$ we get $x=\frac{1-y}{1+y}$.  Then rewrite the floor function as $\lfloor{xy+\frac{1}{xy}}\rfloor=\lfloor{\frac{y^2(1-y)^2+(1+y)^2}{y(1-y^2)}}\rfloor$, so there is no max, taking $y \rightarrow 1$
A: From $xy+x+y=1$ and $x>0$, $y>0$ it follows that $xy<1$. Since $t\mapsto t+{1\over t}$ is decreasing when $t<1$ we conclude that we have to make
 $xy$ is as large as possible. Let $x+y=:s$. Then
$$1-s=xy\leq{s^2\over4}\ .$$ 
The largest possible $xy$ goes with the smallest admissible $s>0$, and the latter satisfies $1-s={\displaystyle{s^2\over4}}$. This leads then to $$x=y={s\over2}=\sqrt{2}-1\ ,$$ and finally to
$$xy+{1\over xy}=6\ ,$$
which is already an integer.
