Prove that $ \lfloor 2x \rfloor \leq 2\lfloor x \rfloor + 1$ 
Theorem. For $x \in \mathbb R$, $$2 \lfloor x \rfloor \leq \lfloor 2x \rfloor \leq 2 \lfloor x \rfloor +1.$$

I tried to prove this theorem by first proving a few helper theorems. I have proved the following Lemma.

Lemma. For $x \in \mathbb R$ and $n \in \mathbb N$, 
  $$n \leq x \Leftrightarrow n \leq \lfloor x \rfloor. $$

This allowed me to prove the first inequality, because 
$$2 \lfloor x \rfloor = \lfloor x \rfloor  + \lfloor x \rfloor \leq x + x = 2x, $$
and because $2 \lfloor x \rfloor$ is integer, $2 \lfloor x \rfloor \leq \lfloor 2x \rfloor$ due to the Lemma above. However, I can not get a hold of the other inequality. Anyone any ideas? Note: I have seen approaches using the fact that $x = \lfloor x \rfloor + x_1$ with $x_1 \in [0,1)$, however I prefer to not take such an approach, and rather work with the properties described above.
 A: $$x = n + p$$
First case $0\leq p < 0.5$:
$$\lfloor 2x \rfloor = 2n \leq 2\lfloor x\rfloor + 1 = 2n + 1$$
Second case $0.5 \leq p < 1$:
$$\lfloor 2x \rfloor = 2n + 1 \leq 2\lfloor x\rfloor + 1 = 2n + 1$$
Proof is completed here.
A: $$\begin{array}{rcl}
x &<& \lfloor x \rfloor + 1 \\
\lfloor 2x \rfloor &\le& 2x \\
\lfloor 2x \rfloor &<& 2\lfloor x \rfloor + 2 \\
\lfloor 2x \rfloor &\le& 2\lfloor x \rfloor + 1 \\
\end{array}$$
A: Let $0\leq\{x\}<\frac{1}{2}.$
Hence, $$[2x]=[2[x]+2\{x\}]=2[x].$$
Let $\frac{1}{2}\leq\{x\}<1$.
Hence, $$[2x]=[2[x]+2\{x\}]=2[x]+1$$
and we are done!
A: Use $\lfloor 2x \rfloor = \lfloor x \rfloor + \lfloor x + 1/2\rfloor$ identity : 
$$\lfloor 2x \rfloor = \lfloor x \rfloor + \lfloor x + 1/2\rfloor \le 2\lfloor x \rfloor  + 1 \to \lfloor x + 1/2\rfloor \le \lfloor x \rfloor  + 1 $$ $$ x  = n + p \ , n\in \mathbb{Z} \ , 0\le p \lt 1$$ 
Then : $$ \lfloor n + p + 1/2 \rfloor \le \lfloor n + p  \rfloor + 1  $$
Using floor function properties : 
$$ n + \lfloor p + 1/2 \rfloor \le n + 1 $$ $$ \lfloor p + 1/2 \rfloor \le 1 $$
Consider two cases ($0\le p \lt 1 \to  1/2\le p + 1/2\lt 3/2 $) : $$ 0\le p+ 1/2 \lt 1 \ (Result : \lfloor p + 1/2 \rfloor  = 0 )$$  and   $$1 \le p+ 1/2 \lt 3/2 \ (Result : \lfloor p + 1/2 \rfloor  = 1 ) $$
In the both cases inequality works . So we proved the inequality for all $x \in \mathbb{R}$ .
A: You will find a property here
https://en.wikipedia.org/wiki/Floor_and_ceiling_functions
that
$$\lfloor u \rfloor + \lfloor v \rfloor 
  \le \lfloor u + v \rfloor
  \le \lfloor u \rfloor + \lfloor v \rfloor + 1$$
letting $u = v = x$, we get
$$ 2\lfloor x \rfloor 
   \le \lfloor 2x \rfloor
   \le 2\lfloor x \rfloor + 1$$
