prove that $\lfloor x\rfloor\lfloor y\rfloor\le\lfloor xy\rfloor$ How can I prove that if $x$ and $y$ are positive then

$$\lfloor x\rfloor\lfloor y\rfloor\le\lfloor xy\rfloor$$

 A: HINT: For this you need only a little more than the fact that if $x\ge 0$, then $0\le\lfloor x\rfloor\le x$ and basic facts about manipulating inequalities. Specifically, you need to realize that if $z\in\Bbb R$, $n\in\Bbb Z$, and $n\le z$, then $n\le\lfloor z\rfloor$.
A: First note that $f:\mathbb R\rightarrow \mathbb Z$ given by $f(x)=\lfloor x\rfloor$ is an increasing function that is the identity on the integers. Then note that for positive $x$ we have $0\leq f(x)\leq x$. With this we get
$$
f(x)f(y)\leq xy
$$
and applying the increasing function $f$ on both sides above noting that the left hand side is an integer we then get:
$$
f(x)f(y)=f(f(x)f(y))\leq f(xy)
$$
which proves the claim.
A: Assuming that you mean $\lfloor x \rfloor \lfloor \color{red}{y} \rfloor \le \lfloor xy \rfloor$,  write
$$x := a + b, \quad y = c + d, \quad a,c \in \mathbb{N} \cup \{0\}, \quad b,d \in [0,1\rangle.$$
Then $\lfloor x \rfloor \lfloor y \rfloor = ac$ and
$$\lfloor xy \rfloor = \lfloor (a+b)(c+d) \rfloor = \dots$$
A: I am using the notation $\{x\}$ to mean the fractional part of $x$.
$$
x = \lfloor x \rfloor + \{x\}, \;\; y = \lfloor y \rfloor + \{y\} \\
xy = \lfloor x\rfloor \lfloor y \rfloor + \lfloor x\rfloor \{y\} + \{x\}\lfloor y\rfloor  + \{x\} \{y\} \\
\lfloor xy\rfloor=\lfloor x\rfloor \lfloor y\rfloor + \lfloor\lfloor x\rfloor\{y\}\rfloor + \lfloor\{x\}\lfloor y\rfloor\rfloor\\
\text{but } x\ge0,\;\;y\ge0\\
\lfloor\lfloor x\rfloor\{y\}\rfloor + \lfloor\{x\}\lfloor y\rfloor\rfloor\ge0\\
\therefore \lfloor xy\rfloor\ge\lfloor x\rfloor\lfloor y\rfloor$$
A: $
\newcommand{floor}[1]{\left\lfloor #1 \right\rfloor}
$Here is how I would write down this proof, essentially the proof from njguliyev's comment.
For all real $\;x,y \ge 0\;$,
\begin{align}
& \floor x \times \floor y \;\le\; \floor{x \times y} \\
\equiv & \qquad \text{"basic property of $\;\floor{\cdot}\;$, using that the left hand side is integer"} \\
& \floor x \times \floor y \;\le\; x \times y \\
\Leftarrow & \qquad \text{"arithmetic: the simplest possible strengthening, using $\;x,y \ge 0\;$"} \\
& \floor x \le x \;\land\; \floor y \le y \\
\equiv & \qquad \text{"basic property of $\;\floor{\cdot}\;$, twice"} \\
& \text{true} \\
\end{align}
A: We can use here the fact that if $x$ and $y$ are positive real numbers, then $[x]\leq\ x$ and $[y]\leq\ y$. So we multiply the inequalities to obtain $[x][y]\leq\ xy$ which implies that $[x][y]\leq\ [xy]$ (by using a well known properties of the  floor function) and this proves the result. (Q.E.D.)
Hope this really helps you! :-)
