A smooth $Q \in \mathbb R^N \to \mathbb R$ close to, but strictly below min

So, I've noticed that in many realworld applications, strict bounds are a requirement. I'll use a factory with $N$ inputs as an example. Suppose the inputs are organised into lots, and one output requires exactly one lot of each input. The factory simply can't run if even one of its inputs runs out, so the factory's maximum production rate (in widgets/second) is

$$P_{max}(I) = \min(I)$$

where $I \in \mathbb R^N$ is a vector of all the input rates.

We'll start with what we're not doing. We aren't worrying about whether the unused inputs will accumulate without bound. In other words, we are supposing that the actual production rate $P$ can be set to any value below or equal to $P_{max}$ without causing issues in the input end. Perhaps the extra is thrown away, or more likely, it is used to inform the future value of $I$, with a low value of $P$ causing it to lower. We're also supposing $I$ is differentiable with respect to time.

What I'm looking at is the problems that arise when you actually do set $P$ to $P_{max}$. Even if the input rates are differentiable with respect to time, $P_{max}$ will not be. Every time the lowest two inputs cross each other, a discontinuity in $P_{max}$'s derivative arises. This produces a discontinuity in the acceleration a.k.a. a jerk of the moving components. In a fake factory like Factorio or Satisfactory, the machines switch on and off instantly, so there isn't even a problem with a discontinuity in velocity—never mind acceleration. In a real factory, the jerk creates vibrations and causes wear, not to mention a lot of noise.

The naïve solution to this problem is to try to set $P$ to some moving average of $P_{max}$, but this is not going to work! Such a moving average carries no guarantee that $P < P_{max}$, which it absolutely must!

A solution that will work is to find some function $Q \in \mathbb R^N \to \mathbb R$ that is smooth and satisfies

$$\forall I, \quad 0 < Q(I) < \min(I)$$

While being as close to $\min(I)$ as possible within the limits of the maximum mechanical jerk.

I've been taking some stabs at some candidates, and there are some very good solutions for the $N = 2$ case, based off of smoothings of the absolute or sign functions, but nothing immediately good and simple comes to mind for the general case. Do any of you know a function like this?

While researching this problem I actually found an answer! In real analysis, there's a much-used result:

$$\min(I) = \lim_{k \to \infty} \left( \sum_{a = 0}^{<N} \frac 1 {I_a^k} \right)^{- \frac 1 k}$$

To get the $Q$ I'm looking for, all I have to do is get rid of the limit:

$$Q(I) = \left( \sum_{a = 0}^{<N} \frac 1 {I_a^k} \right)^{- \frac 1 k}$$

This function is smooth and strictly smaller than the minimum.

To choose $k$, the easiest thing to do is substitute in the "even keel" state, where all input rates are equal. Say $\forall a, I_a = v$. If you require that in this state, $Q(I) = \alpha v$ for some $0 < \alpha < 1$, then

$$k = -\frac {\log N} {\log \alpha}$$

Thus to maintain $\alpha$, $k$ only needs slight adjustment, even for large increases in $N$.