$2 \lfloor x \rfloor \leq \lfloor 2x \rfloor \leq 2 \lfloor x \rfloor +1$ May I know the standard proof technique to prove such kind of inequalities.
$2 \lfloor x \rfloor \leq \lfloor 2x \rfloor \leq 2 \lfloor x \rfloor +1$
Thanks!

By Hermite's identity, we know that $ \lfloor x \rfloor + \lfloor x \rfloor \le \lfloor x \rfloor + \lfloor x + \frac 12 \rfloor = \lfloor 2x \rfloor \le \lfloor x \rfloor + \lfloor x + 1 \rfloor$. Alternatively, as already mentioned, you can use casework on $\{x\} := x - \lfloor x \rfloor$, in particular when $0 \le \{x\} < 1/2$ and when $1/2 \le \{x\} < 1$. 
