$2 \lfloor x \rfloor \leq \lfloor 2x \rfloor \leq 2 \lfloor x \rfloor +1$

Question

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!

Answer 1

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 $1/2 \le \{x\} < 1$.

Answer 2

Hint: let $n = \lfloor{x\rfloor}$, so $n \le x < n+1$. What about $2x$?