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
 A: I shall assume that $x, y \ge 0$. We also have $\lfloor x\rfloor, \lfloor y\rfloor \ge 0.$
Write $x = \lfloor x\rfloor + \{x\}$ and $y = \lfloor y\rfloor + \{y\}.$ Clearly, $0 \le \{x\}, \{y\} < 1$.  
Note that
$$xy  =\lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor\{y\} + \lfloor y\rfloor\{x\} + \{x\}\{y\}. \quad (*)$$

Using the fact that all the rightmost three terms on the RHS of $(*)$ are nonnegative, we see that
$$xy \ge \lfloor x\rfloor\lfloor y\rfloor.$$
Note that the RHS is an integer which is lesser than $xy$. By definition of floor, $\lfloor xy\rfloor$ must be the greatest such integer. Thus, we have
$$\lfloor xy\rfloor \ge \lfloor x\rfloor\lfloor y\rfloor,$$
giving us the left inequality.

Using $(*)$ and the fact that $\{x\}, \{y\} < 1$, we see that
$$xy \le \lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y\rfloor + \{x\}\{y\}.$$
Since $\lfloor .\rfloor$ is an increasing function, we see that
$$\lfloor xy\rfloor \le \lfloor\lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y \rfloor+ \{x\}\{y\}\rfloor. \quad (**)$$
Note that $\lfloor n + z\rfloor = n + \lfloor z\rfloor$ for any $n \in \Bbb Z$ and $z \in \Bbb R$. Using this, the RHS of $(**)$ can be written as
\begin{align}
&\lfloor\lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y \rfloor+ \{x\}\{y\}\rfloor\\
=&\lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y \rfloor+ \lfloor\{x\}\{y\}\rfloor\\
=& \lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y \rfloor, \quad (\because 0 \le \{x\}\{y\} < 1)
\end{align}
giving us the right inequality.
A: 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.
