Odd square roots and the sum of element orders on Sn and An

For a finite group $G$ let $\psi (G)$ be the sum of the orders of its elements. The quotient $S_n\mathrm{/}{A}_n$ has order two. Hence $\psi (S_n\mathrm{/}{A}_n)=3$. The quantity to be compared with $\psi (S_n)$ is therefore $3\psi (A_n)$. Computation indicates that $3\psi (A_n)>\psi (S_n)$ for every $n\ge 3$. We reduce this inequality to a question on the parity of square roots. Let $r(\beta )$ be the number of odd permutations $\sigma$ with ${\sigma }^2=\beta$. We show that the sum of the orders of the odd permutations equals $2{\sum }_{\beta \in A_n}r(\beta )o(\beta )$. We also show that ${\sum }_{\beta \in A_n}r(\beta )$ equals $\mathrm{\mid }{A}_n\mathrm{\mid }$. The inequality then says that $r$ and the order function are negatively correlated on $A_n$. It says equally that the average order of an odd permutation is less than twice the average order of an even one. We then rule out two natural approaches. No injection from the odd permutations into $A_n$ can halve the order at every point. The smallest failure occurs at $n=12$. A threshold argument through Landau’s function fails as well. Computations up to $n=60$ are reported.