I want to see if for all $n, k \in \mathbb{Z}, k \geq 2$, $$\left\lfloor{\frac{n}{k}}\right\rfloor \leq \left\lceil{\frac{n+1}{2}}\right\rceil$$ I've checked a bunch of numbers with some basic python, but how would I show this mathematically?
Edit: I have a lot more inequalities like this to show, so I'd love to also know the method you would use. Thanks!
The statement does not hold... Even if $n,k \in\mathbb{N}$, in the case $k=1$ you would have to show that $$ n \leq \left\lceil \frac{n+1}{2} \right\rceil $$
and this is just nor true. If $n,k \ge 2$ the result holds. To check that, just need to show that it holds for $k=2$, because for larger $k$ the LHS is even smaller.
Considering any $k \ge 2$, the result does not hold for in general for negative $n$. For instance, if $k\ge 4$, the result only holds for $n\ge -2$.