Prove that $⌊xy⌋ \le ⌊x⌋⌊y⌋ + ⌊x⌋+⌊y⌋$ for $x,y \in \mathbb{R}$, as $x,y>0$

Question

I'm trying to prove that $⌊xy⌋ \le ⌊x⌋⌊y⌋ + ⌊x⌋+⌊y⌋$ by writing x as $x=m+\epsilon_1$ and $y=n+\epsilon_2$ for some positive integer $m,n$. Trying to use the inequality $⌊xy⌋ \le xy$ and $(m+\epsilon_1)(n+\epsilon_2)<m+n+mn$, but it turns out that the inequality doesn't hold, and I have no idea how to prove this inequality then.

Can somebody share some lights on this issue?

Thanks!

Answer 1

Let $x = m + a$ and $y = n + b$, where $m,n \in \mathbb{Z}$ and $a,b \in [0,1)$. So, the problem comes down to $$\lfloor (m+a)(n+b) \rfloor \le mn + m + n$$ Since $ab \in [0,1)$, we have $$ \lfloor mn + an+bm + ab \rfloor \le \lfloor mn + m + n +ab \rfloor \tag 1 $$ Again since $ab \in [0,1)$ and $mn + m +n \in \mathbb{Z}$, we have $$\lfloor mn + m + n +ab \rfloor =mn + m + n \tag 2$$ Now, $(1)\land (2)$ implies $$\lfloor (m+a)(n+b) \rfloor \le mn + m + n$$ or $$\lfloor xy\rfloor \le \lfloor x \rfloor \lfloor y \rfloor + \lfloor x \rfloor +\lfloor y\rfloor $$ as desired.

Answer 2

We have $$(m+\epsilon_1)(n+\epsilon_2)<(m+1)(n+1)=mn+m+n+1$$Now apply the floor function to the left-hand side. Knowing that $mn+m+n+1$ is an integer gives you your result.

Answer 3

Instead of $x$ I am writing $X+x$ where $X$ is the integer part ($X\ge0,0\le x<1$).

We have

$$\lfloor (X+x)(Y+y)\rfloor\le XY+X+Y \iff\lfloor(x-1)Y+(y-1)X+xy\rfloor\le 0$$

and

$$\lfloor(x-1)Y+(y-1)X+xy\rfloor\le\lfloor xy\rfloor=0.$$