Let $x \in [0, 1]^N$, and $y \in [0, 1]$. Consider also $a >0$ and $b > 0$.
Define the following linear function:
$$f_s(x,y) = \sum_{i=1}^N[(a-b)x_i-a] - s[(a-b)y - a],$$
for $s>0$.
What is the minimum $s$ such that $f_s(x,y)$ is non-negative for all $x \in [0, 1]^N$ and for all $y \in [0,1]$?
My attempt
It is easy to show that:
$$\min\{-a, -b\} \leq (a-b)x_i-a \leq \max\{-a, -b\},$$
or equivalently
$$-\max\{a, b\} \leq (a-b)x_i-a \leq -\min\{a, b\}.$$
Of course, the same holds replacing $x_i$ with $y$.
Then:
$$\begin{align}f_s(x,y) & \geq \sum_{i=1}^N(-\max\{a, b\}) - s(-\min\{a, b\}) = \\ & = -N\max\{a, b\} + s \min\{a, b\}. \end{align} $$
In order to guarantee that $f_s(x,y) \geq 0$, then, I conclude that $$s \geq N\frac{\max\{a, b\}}{\min\{a, b\}}.$$
This bound works. Anyway, from numerical simulation I observed that there can be a better lower bound (I tried generating random numbers for $10^6$ times).
Is there another way to find the minimum $s$ possible?
Remark
The bound I found is the best when $a=b$.