Given a positive integer $n$, recall that that the value of the divisor function $\sigma$ at $n$ is defined to be the sum of all the divisors of $n$; in symbols: $$ \sigma(n) = \sum_{d \mid n} d. $$
Is there an efficient way to determine the maximum of $\sigma(n)/n$ for $n \in [a,b] \cap \Bbb{Z}$?
Here by efficient I mean anything faster than actually computing $\sigma(n)/n$ for all $n$ in the interval.
You can say some things, but perhaps not in any great detail; the "high water marks" of the expression $\sigma(n)/n$ occur at the superabundant numbers, so in any interval of the form $[a,b]$ the ratio will be maximized by the largest superabundant number in that interval (provided one exists).
The bad news is that there is not a great deal of control over the prime factorizations of the superabundant numbers, though perhaps you will be able to get enough of a speedup with what little is easily known.