A proof regarding floor I have been trying to prove that:
$$x \geqslant 1 \implies \text{floor}(x) \geqslant \frac{x}{2}$$ 
I know this claim is true because I have compared the graph of $x/2$ and $\text{floor}(x)$. However, I just can't seem to get the right intuition has to how to go about proving this. I'm not asking for the proof. I'm asking for a hint, or maybe something that will push me in the right direction.
 A: Hint:
$$\lfloor x \rfloor > x-1.$$
A: If $x\ge 1$ then $\lfloor x\rfloor \ge 1$
Let $1\le \lfloor x\rfloor =k\le x < k+1$
Then $ \frac x2 < \frac {k+1}2=\frac k2 + \frac 12\le \frac k2+\frac k2=k=\lfloor x\rfloor $
A: Either $\lfloor x \rfloor = 2k$ or $\lfloor x \rfloor = 2k+1$ for some integer $k \ge 0$.
CASE $1. \quad$ $x = 2k + \epsilon$ where $0 \le \epsilon < 1$
\begin{align}
   \lfloor x \rfloor \ge \dfrac x2
   &\iff 2k \ge k + \dfrac{\epsilon}{2} \\
   &\iff k \ge \dfrac{\epsilon}{2} \\
   &\iff \text{$(k=\epsilon = 0)$ or $(k \ge 1)$} \\
   &\iff \text{$(x=0)$ or
         $(x = 2k + \epsilon$ where $k \ge 1$ and $0 \le \epsilon < 1)$}
\end{align}
CASE $2. \quad$ $x = 2k + 1 + \epsilon$ where $0 \le \epsilon < 1$
Note that $0 \le \epsilon < 1 
           \iff  -\dfrac 12 \le \dfrac{\epsilon}{2} - \dfrac 12 < 0$
\begin{align}
   \lfloor x \rfloor \ge \dfrac x2
   &\iff 2k + 1 \ge k + \dfrac 12 + \dfrac{\epsilon}{2} \\
   &\iff k \ge \dfrac{\epsilon}{2} - \dfrac 12 \\
   &\iff \text{$(x = 2k + 1 + \epsilon$ 
                where $k \ge 0$ and $0 \le \epsilon < 1)$}
\end{align}

So $ \lfloor x \rfloor \ge \dfrac x2$ iff $x=0$ or $ \lfloor x \rfloor \ge 1$
