Prove that $2\lfloor y \rfloor > y$ when $y \geq 1$

Problem:

Prove that $2\lfloor y \rfloor > y$ when $y \geq 1$.

Proof:

Let $y = \lfloor y \rfloor + r$, then $0 \leq r < 1$.

Besides, since $y \geq 1$, we have $\lfloor y \rfloor \geq 1$, and thus $\frac{1}{\lfloor y \rfloor} \leq 1$.

From $r < 1$,

$$\frac{r}{\lfloor y \rfloor} < \frac{1}{\lfloor y \rfloor} \leq 1 \\ \frac{r}{\lfloor y \rfloor} + 1 = \frac{r + \lfloor y \rfloor}{\lfloor y \rfloor} < 1+1 \\ \frac{y}{\lfloor y \rfloor} < 2 \\ y < 2 \lfloor y \rfloor \\ 2 \lfloor y \rfloor > y \quad \blacksquare$$