find $u < r$, $v < s$ such that $p< uv < rs$ Let $p < rs$ where both $p, r, s$ are positive rational numbers.
I want to find rationals $u < r$, $v<s$ such than $p < uv < rs$ holds.
It seems obvious to me since for sufficient small $\epsilon$, $u = r - \epsilon$, $v = s - \epsilon$ is my desired.
But how can I construct such $\epsilon$ exactly?
 A: Addendum added to correct an analytical error in my answer.

Without loss of generality, $u,v$ are both positive.  That is, since $0 < p < uv,$ then $u$ and $v$ must have the same sign.  This means that $(|u|,|v|)$ is a satisfying ordered pair if and only if $(-|u|,-|v|)$ is also a satisfying ordered pair.

Edit 
Actually, the above paragraph contains an analytical error.  See the Addendum.

Further, with $u,v$ both assumed positive, 
$u < r, v < s$ automatically implies $uv < rs.$
So, the question reduces to finding positive $u,v$ such that

*

*$p < uv.$

*$u < r, v < s.$
One way of doing this is to establish a range of values of $u$ and then, for each value of $u$, establishing a range of values for $v$.
The easiest way of doing this is to temporarily pretend that the constraints are:

*

*$p \leq uv.$

*$u \leq r, v \leq s.$
So, under the pretend constraints above, the largest value for $u$ is $r$.  This implies that the smallest value for $v$ is given by $\dfrac{p}{r}.$
Similarly, the largest value for $v$ is $s$, and the smallest value for $u$ is given by $\dfrac{p}{s}.$
So, you have that $\dfrac{p}{s} \leq u \leq r.$
Then, you have that for any value of $u$, you must have that
$\dfrac{p}{u} \leq v \leq s.$
Putting this all together, and reverting back to the actual constraints, the complete set of all (positive) $(u,v)$ is given by
$$\{ ~(u,v) ~: ~ \frac{p}{s} < u < r, ~\frac{p}{u} < v < s ~\}.$$

Addendum
It is true that if $(|u|,|v|)$ is a satisfying ordered pair, then so is $(-|u|, -|v|).$
However, the converse is false.
For example, take $r,s$ = 2,3, respectively.
Take $p = 1.$
Then $\displaystyle ~(u,v) = \left( ~-10^6, ~-\frac{1}{10^6 - 1} ~\right)$ is a satisfying ordered pair.
So, when considering satisfying ordered pairs $(u,v)$, where $0 > u,v$, the only relevant constraints are that $p < uv < rs.$
So, $u$ can be any negative number.
Then, the set of all satisfying (negative) values of $v$ is given by
$$\frac{p}{|u|} < |v| < \frac{rs}{|u|} \implies \frac{p}{u} > v > \frac{rs}{u}.$$
A: *

*Find any $k_1$ satisfying $\quad 2^{k_1} > \frac{r+s}{rs-p}\qquad (1)$

*Find any $k_2$ satisfying $\quad 2^{k_2} > \frac{1}{r+s}\qquad (2)$

*Let $k = \max\{k_1,k_2\}.\ $ Then $\ k\ $ (in place of $k_1$ and $k_2$) satisfies both $(1)$ and $(2).$

*Then $u = r - 2^{-k},\ v = s - 2^{-k}\ $ will do, since:

$(i)\quad(1)\implies rs - (r+s)2^{-k} > p,\implies uv=rs - (r+s)2^{-k} + 2^{-2k} > p,\ $ and
$(ii)\quad (2) \implies r+s > 2^{-k}, \implies 2^{-k}(r+s) > 2^{-2k},\implies 2^{-k}(r+s) - 2^{-2k} > 0,\implies rs - (2^{-k}(r+s) - 2^{-2k}) < rs.$
A: The $\epsilon$ requested by the OP has to satisfy 2 conditions.
The first condition is:
$$rs-(r-\epsilon)(s-\epsilon) \gt 0$$
$$\epsilon(r+s-\epsilon) \gt 0$$
Clearly any $\epsilon$ satisfying
$$0 \lt \epsilon \leq \frac{r+s}{2} \tag{1}$$
meets the requirement.
The 2nd condition is
$$(r-\epsilon)(s-\epsilon)-p \gt 0$$
$$rs-p - \epsilon(r+s)+ \epsilon^2 \gt 0$$
This condition is satisfied when
$$0 \lt \epsilon \leq \frac{rs-p}{2(r+s)}   \tag{2}$$
Combining (1), (2), we may let
$$\epsilon = \min \left\{ \frac{r+s}{2}, \frac{rs-p}{2(r+s)} \right\}$$
A: In this answer, we will construct a closed form example for the rational pair $(u,v)$.

Your exact problem is:

I want to find rationals $u < r$, $v<s$ such than $p < uv < rs$ holds.

Let $u,v>0$ and $\alpha,\beta >1$, then our first move is to set up the following system of equations:
$$\begin{align}\begin{cases}u=\dfrac r\alpha \wedge v=\dfrac s\beta \\
uv=\dfrac {p+rs}{2}\end{cases}\end{align}$$
This implies that,
$$\begin{align}&\frac {rs}{\alpha\beta}=\frac {p+rs}{2}\\
\implies &\alpha\beta=\frac {2rs}{p+rs}\\
\implies &\alpha=\frac {2rs}{\beta({p+rs})}\end{align}$$
where $1<\beta<\frac {2rs}{p+rs}$, since $2rs>p+rs$ and $\alpha>1$.
Therefore, we can take:
$$\begin{align}\beta=\frac{1+\frac {2rs}{p+rs}}{2}=\frac{p+3rs}{2(p+rs)}\end{align}$$
This leads to:
$$\begin{align}\alpha=\frac {4rs}{p+3rs},~\beta=\frac{p+3rs}{2(p+rs)}\end{align}$$
Finally, the closed form example we want to construct is:
$$\begin{align}\bbox[5px,border:2px solid #C0A000]{u=\frac {p+3rs}{4s},~v=\frac {2s(p+rs)}{p+3rs}}\end{align}$$
