Let $G$ be a finite group of order $n$ and $\psi(G)$ be the sum of element orders of $G$. Then $\psi(G)\leq\psi(C\_n)$, where $C\_n$ is the cyclic group of order $n$ (see "Sums of element orders in finite groups", Comm. Algebra 37 (2009), 2978-2980). Is it true a similar inequality for the product of element orders of $G$? 



A question on the product of element orders of a finite group
Denoting the order of $g$ by $o(g)$, you can show that for any decreasing function $f$ the following inequality holds
$$\sum\_{g\in G}f(o(g))\geq \sum\_{g\in \mathbb Z/n\mathbb Z}f(o(g)).$$
This is because one can actually construct a bijection $\sigma:G\to\mathbb Z/n\mathbb Z$ which satisfies $$o(\sigma(g))\geq o(g)$$ for all $g\in G$. The main ingredient is a classical [theorem of Frobenius](http://www.pitt.edu/~gmc/ch1/node7.html) saying that when $k$ divides the order of a group, the number of elements of order dividing $k$ is divisible by $k$, then proceed by induction. An application of this exact idea is for example [problem 10775](http://www.jstor.org/stable/2695368) on the American Math Monthly. For your question we just need $f(x)=-\log x$.


