If Robin's inequality ever fails, are there only finitely many colossally abundant numbers that satisfy it? Let$\ \sigma(n)$ be the sum-of divisors function, with the divisors raised to$\ 1$. If the Riemann Hypothesis is false, Robin proved there are infinitely many counterexamples to the inequality$$\ \sigma(n)<e^\gamma n \log \log n.$$ There are 27 small counterexamples, but the conjecture is that it holds for every$\ n>5040$. Akbary and Friggstad showed the least counterexample to it must be a superabundant number, i.e. a number$\ a$ such that$\ \frac{\sigma(a)}{a}>\frac{\sigma(b)}{b}$ for all$\ b<a$. Now, it is a virtual certainty that if the inequality fails (for some$\ n>5040$), the maximum of the ratio$\ \frac{\sigma(n)}{n \log \log n}$ will be reached by a colossally abundant number, namely a number$\ c$ such that$\ \frac{\sigma(c)}{c^{1+\epsilon}}>\frac{\sigma(d)}{d^{1+\epsilon}}$, for all$\ d<c$ and for some$\ \epsilon>0$. Since it could lead me to something on the subject, what I'm asking is: if the inequality fails, will only a finite number of colossally abundant numbers satisfy Robin's inequality?
 A: For the benefit of those who may not be familiar with all this:
In 1915 Ramanujan proved that if the Riemman Hypothesis is true, then for all sufficiently large $n$ we have an inequality on $\frac{\sigma(n)}{n}$, where $\sigma(n)$ is the sum of the divisors of the positive integer $n$. The inequality was $$\sigma(n)<e^\gamma n \ln\ln n$$ In 1984 Robin elaborated this to show that if there is a single exception to this for $n>5040$ (the largest currently known exception and a "colossally abundant number" - hereafter a "CA"), then the Riemann Hypothesis is false.
Because of the importance of the RH, this attracted a good deal of attention. But 30 years later nothing seems to have come of it.
Obviously the most plausible candidates to break the inequality are numbers with lots of divisors. I believe, though I am weak on the history, that the concept, if not the name, of CA came from Ramanujan during his 1915 work.
To give a little perspective: few people are interested in CA per se. But vast numbers of people are interested in RH, even if only a tiny number do serious work on it (because of the risk to one's reputation). So the immediate interest of the inequality was that it provided another way, superficially at least totally different, to disprove the RH by computation. People had got fed up with results that the first zillion zeros were on the line, particularly when analysts quoted Littlewood's "Miscellany" on the Skewes' number (which is now a somewhat less compelling point :) ). So this was something else to try.
However, after 30 years nothing has so far come of that. In the meantime people have been working on CA as objects of interest in their own right.
The question is whether if the RH is false (so that the inequality fails - Robin's result was an iff type result), then only a finite number of CA will satisfy the Robin inequality.
[Added later - the precise question having been clarified]
If I had realised that would be the question, I would never have started to answer it! I had earlier understood it to be a quite different question. But there are a few points to be made.
I have never read Robin's paper - my interest is in RH, and I do not regard Robin's inequality as a useful way of tackling the RH (a judgment which of course is of zero interest to anyone else). So I am at a serious disadvantage - in not having read the paper and to compound that, I cannot immediately lay my hands on it.
It is fairly easy to show that if $a,b$ are coprime counterexamples to the inequality, then so is $ab$ provided $a,b$ are sufficiently big (which they would be). It is also fairly clear that unless something weird happens at huge values, counterexamples are likely to be CA. So it seems a fairly safe guess that if RH is false then there will be infinitely many CA not satisfying the Robin inequality. 
But unfortunately the question asks for something much stronger than that, namely will all but finitely many CA fail to satisfy it?
Short answer: good question; I have no idea and should delete this entire answer. But pending a little digging early in the coming week I will leave it here until a better answer comes.
In my defence, I would only say that I have only been using this site for less than 3 weeks. I have answered lots of daft questions, and had fun competing putting up answers fast. I failed to adjust adequately when this one came along. But it does illustrate the wisdom of the concept of clarifying the question with comments before writing Answers. I had started to do that, but got impatient when I could not immediately grasp the clarifications. That was entirely my fault. I apologise unreservedly.
A: I just stumbled across this question many years after the posting and did not see what I believe to be the answer. If RIE fails the only requirement on colossally abundant numbers being exceptions is that there must be at least one.
There could be only one c.a. exception. Or there could be a finite number of c.a. exceptions. Or there could be an infinite number of c.a exceptions.
If there are only a finite number of ca exceptions, then that would require there be an infinite number of colossally abundant numbers that satisfy RIE.
If there are an infinite number of ca exceptions then there may be a finite or an infinite number of colossally abundant numbers that satisfy RIE.
