Bound on geometric mean of gcd

Question

Given positive integer number $n$, based on Bounds on a sum of gcd's, we know that "Arithmetic mean" is unbounded $$\frac{\sum_{i=1}^n\sum_{j=1}^n\gcd (i, j)}{n^2}=\frac{6}{\pi^2}\ln(n)+O(1).$$ This led me to contemplate "Geometric mean", by using computer I found that $$\frac{\sum_{i=1}^n\sum_{j=1}^n\ln\big(\gcd (i, j)\big)}{n^2}<6\iff \sqrt[n^2]{\prod_{i=1}^n\prod_{j=1}^n\gcd (i, j)}<e^6.$$ How can this be proven?

Answer 1

Whenever you're dealing with the gcd function within a summation, it's usually easier to first sum over the values of the gcd, then generate the numbers in terms of the gcd. In this case, $d$ denotes the gcd of $i$ and $j$, and the sum on the right shows how you'd generate the values of $j$ in terms of $d$.

$$\sum_{i=1}^n{\sum_{j=1}^n{\ln(\gcd(i, j))}} = \sum_{i=1}^n{\sum_{d|i}{\ln(d)\sum_{\ \ \ j=1\\ (i,j)=d}^n{1}}}$$

In this case, the condition $(i,j)=d$ is equivalent to $j$ being a multiple of $d$ where the mutliple is coprime to $i$. This is so that $i$ and $j$ don't have a larger common divisor than $d$. We get

$$\sum_{i=1}^n{\sum_{d|i}{\ln(d)\sum_{\ \ dk \le n\\ (k, i)=1}{1}}}\ \lt\ \sum_{i=1}^n{\sum_{d|i}{\ln(d)\frac{n}{d}}}$$

Rearranging leads to

$$n\sum_{d\le n}{\frac{\ln(d)}{d}}\sum_{i\le n\\d|i}{1}\ \le n\sum_{d\le n}{\frac{\ln(d)}{d}\cdot\frac{n}{d}}$$

which is clearly bounded by

$$n^2\sum_{d\ge 1}{\frac{\ln(d)}{d^2}} \approx 0.937548\cdot n^2$$

This means the geometric mean is bounded above by $e$ to this value which is around $2.5537$.