Proof $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le \lfloor{x + y}\rfloor $

Question

I want to prove that $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le \lfloor{x + y}\rfloor $, I started proving but I got stuck.

Let $ x, y \in \mathbb{R} $. Therefore:

$ \lfloor{x}\rfloor \le x $

$ \lfloor{y}\rfloor \le y $

$ \Rightarrow $

$ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le x + y $

Answer 1

Is this proof OK, using both (1) integers $n = \left\lfloor n \right\rfloor$, and (2) if $a\le b$ then $\left\lfloor a\right\rfloor \le \left\lfloor b \right\rfloor$? $$\left\lfloor x\right\rfloor + \left\lfloor y \right\rfloor = \big\lfloor \left\lfloor x\right\rfloor + \left\lfloor y \right\rfloor \big\rfloor \le \left\lfloor x + y \right\rfloor$$

Answer 2

$ \lfloor{x}\rfloor $ is an integer $ \leq x;\ \lfloor{y}\rfloor $ is an integer $ \leq y.$

Therefore, $ \lfloor{x}\rfloor + \lfloor{y}\rfloor $ is an integer that is $ \leq x+y.$

Since $\lfloor{x+y}\rfloor$ is the greatest integer $ \leq x+y,$ and $ \lfloor{x}\rfloor + \lfloor{y}\rfloor $ is an integer that is $ \leq x+y,$ it follows that $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \leq \lfloor{x+y}\rfloor.$