Proving that there exists integers such that at least one of three inequalities are satisfied Suppose I am given 4 positive integers $A, B, C, D$, where $A > B$ and $C < \lceil \frac{A}{B} \rceil $, how can I prove that there always exists integers $x,y \geq 0$, where $D = x+y$ such that at least one of these three inequalities must be satisfied for every valid combination of $A, B, C, D \in \mathbb{Z}_{> 0}$.
$$\frac{y}{x+C} \geq \frac{A}{B} \geq \frac{y-1}{x+1}$$
$$\frac{y+C}{x-1} \geq \frac{A}{B} \geq \frac{y+1}{x}$$
$$\frac{y+C}{x} \geq \frac{A}{B} \geq \frac{y}{x+1}$$
I did some algebraic manipulation to arrive at the above three inequalities in that form, but can't seem to reason anything about the existence of such an $x, y$. Simulations seem to suggest that there always exists such an $x, y$ (though not exhaustive!), but I can't find a proof for it.
Does anyone have any insights on how to address this? (Or perhaps a counterexample otherwise).

Replace $\frac AB$ with $r-1$.
Thus, given $C,D\in\Bbb Z_{>0}$ and a real number $r>C$, the claim is that one of
$$\frac {D-x}{x+C}\ge r-1\ge \frac {D-x-1}{x+1} $$
$$\frac {D-x+C}{x-1}\ge r-1\ge \frac {D-x+1}{x} $$
$$\frac {D-x+C}{x}\ge r-1\ge \frac {D-x}{x+1} $$
has an integer solution $0\le x\le D$ (with $x=0$, $x=1$ excluded for the lower two options).
Equivalently, one of
$$\frac {D+C}{x+C}\ge r\ge \frac {D}{x+1} $$
$$\frac {D+C-1}{x-1}\ge r\ge \frac {D+1}{x} $$
$$\frac {D+C}{x}\ge r\ge \frac {D+1}{x+1} $$
has such a solution.  Take reciprocals,
$$\frac {x+C}{D+C}\le \frac1r\le \frac {x+1}D $$
$$\frac {x-1}{D+C-1}\le \frac1r\le \frac {x}{D+1} $$
$$\frac {x}{D+C}\le \frac1r\le \frac {x+1}{D+1} $$
and solve for $x$,
$$ \frac{D+C}r-C \ge x \ge \frac Dr-1 $$
$$ \frac{D+C-1}r+1 \ge x \ge \frac {D+1}r $$
$$ \frac{D+C}r \ge x \ge \frac {D+1}r-1 $$
The differences between upper and lower bound are
$$ \frac Dr+1-C,\qquad \frac{C-2}r+1,\qquad \frac{C-1}r+1$$
As $\frac{C-1}r+1\ge1$, there is always an integer $x$ satisfying $\frac{D+C}r \ge x \ge \frac {D+1}r-1$. As $r>1$, we have $\frac{D+1}r-1<D$ and hence the smallest integer in that range is indeed $\le D$, and as $\frac {D+C}r>0$, the greatest such integer is $\ge 0$. So it seems already the third option always has a solution, but for that, we must require $x\ne 0$. In other words, this works only if $$\tag1\frac{D+C}r\ge 1.$$
On the other hand, if $\frac{D+C}r<1$, then one sees that the second inequality has at most $x=1$ as solution - which is not allowed, and the first inequality cannot have a non-negative solution at all. Thus $(1)$, or equivalently
$$ {B(D+C-1)}\ge A$$
is necessary.
