Proving an inequality involving floor division of positive integers?

Question

Let $a$, $b$ and $c$ be positive integers.

Is it always true that $a > \lfloor c / b \rfloor$ implies $ab > c$?

If it is not true, what is a counterexample?

If it is always true, how can we prove it?

Answer 1

With the help of @peterwhy's comment I could work out a solution.

Since $a$ and $\lfloor c/b \rfloor$ are integers, $a \ge \lfloor c/b \rfloor + 1 = c/b - (c/b - \lfloor c/b) \rfloor + 1 = c/b + (1 - (c/b - \lfloor c/b \rfloor))$.

Since $0 \le (c/b - \lfloor c/b \rfloor) < 1$, we can say $0 < (1 - (c/b - \lfloor c/b \rfloor)) \le 1$.

So $a \ge c/b + (1 - (c/b - \lfloor c/b \rfloor)) > c/b .$

Thanks for the comments!

Answer 2

Approach that I prefer:

Since $~c,b~$ are positive integers, there exists $~P,r~$ such that
$P \in \Bbb{Z}, ~0 \leq r < b,~$ and $~c = Pb + r.$

This implies that $~\displaystyle P = \left\lfloor\frac{c}{b}\right\rfloor \implies a \geq (P+1).~$

Therefore

$\displaystyle ab \geq (P+1)b = Pb + b > Pb + r = c.$