Find $\min\big\{ \lfloor xy + \frac{1}{xy} \rfloor \,\Big|\, (x+1)(y+1)=2 ,\, 0<x,y \in \mathbb{R} \big\}$

Question

I 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)

Answer 1

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.

Answer 2

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

1. $ xy + x+ y = 1 $
2. $ 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.

Answer 3

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 $y \rightarrow 1$

Answer 4

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 $x y+\frac{1}{x y}\geq 6$