I want to show that for any nonnegative integers $l$ and $b$ we have $$ \frac{l}{2^{b+1}} - 1 \leq \left\lfloor \frac{l-1}{2^{b+1}} \right\rfloor. $$
I have a proof where I wrote $l = \alpha\cdot 2^{b+1} + \beta$ with $0\leq \beta\leq 2^{b+1}$, but it's not very elegant, I say.
Let $l = \alpha\cdot 2^{b+1} + \beta$ with $0\leq \beta\leq 2^{b+1}.$ Then we have (LHS): $$ \frac{l}{2^{b+1}}-1 = \alpha + \frac{\beta}{2^{b+1}}-1 $$ and (RHS): $$ \left\lfloor \frac{l-1}{2^{b+1}}\right\rfloor = \left\lfloor \alpha + \frac{\beta-1}{2^{b+1}}\right\rfloor. $$
In case $\beta = 0$ we get $\alpha - 1$ for (LHS) and due to $0<2^{-(b+1)}<1$ we get the same for (RHS), so we have equality for this case.
Otherwise, we have $1\leq \beta \leq 2^{b+1}$ and so $0 \leq \frac{\beta - 1}{2^{b+1}} < \frac{\beta}{2^{b+1}}<1$. So we have for (LHS) $$ \frac{l}{2^{b+1}}-1 = \alpha + \frac{\beta}{2^{b+1}}-1<\alpha $$ and for (RHS) $$ \left\lfloor \frac{l-1}{2^{b+1}} \right\rfloor = \left\lfloor \alpha + \frac{\beta-1}{2^{b+1}} \right\rfloor = \alpha.\square $$
Is there a more elegant way of showing this?