Floor function of a product I'm reading a book about proofs and I'm currently stuck in this problem.
Prove that for all real numbers $x$ and $y$ we have that:
$$\lfloor x\rfloor \lfloor y\rfloor \leq \lfloor xy\rfloor \leq \lfloor x\rfloor \lfloor y \rfloor + \lfloor x \rfloor + \lfloor y \rfloor$$
I though I could do it by cases, considering when the number is a pure integer and when is an integer plus some real number. But by doing this I end up with having a lot of cases to show. Is there any better, simpler and clever approach? Thank you!
PS: I end up with a lot of cases because there is a point where I will have to consider "subcases" of cases specifically when the integer part is multiplying with the positive "rest" less than one

It can help to write $x=a+r$ where $a$ is an integer and $0\leq r <1$, so that $\lfloor x \rfloor = a.$
Similarly, let $y=b+s$ where $b$ is an integer and $0\leq s <1.$
Then $$\lfloor x \rfloor \lfloor y \rfloor = ab$$  and
$$\lfloor x y \rfloor = \lfloor ab+as + br +sr \rfloor  = ab +\lfloor as + br +sr \rfloor.$$
(Assuming $x$ and $y$ are positive,) the last term above has to be greater than $0$, so that's your first inequality.
Then take that last term, and since $r$ and $s$ are less than one:
$$\lfloor as + br +sr \rfloor \leq \lfloor a + b +sr \rfloor = a + b + \lfloor sr \rfloor.$$
The last term is $0$, so there's your second inequality.
