I believe that the sum of the lengths of diagonals of a regular polygon ($n$-gon) is always greater than or equal to any other irregular polygon ($n$-gon) inscribed in a circle.
For example for a 4-gon, if we consider a regular $4$-gon, then the two diagonals pass through the centre of a circle, resulting in the sum of the diagonals' lengths twice the diameter. Considering any other $4$-gon (irregular), the sum of the lengths of the diagonals is not greater than $2\times$ the diameter.
Case : 5-gon I proved that the sum of the lengths of diagonals in a regular $5$-gon is maximum compared to any $5$-gon (irregular polygon). I used Lagrange's multiplier to prove it.
Case : $n$-gon I tried proving the same for an $n$-gon using Lagrange's multiplier, where we need to maximize the sum of the lengths of all diagonals with respect to $$\theta_1+ \theta_2 + \dots + \theta_n =2\pi,$$ but unable to do the calculation as the number of variables increases. Here $\theta_i$ is the angle subtended to the centre of a circle by the side $a_i$ of the polygon.
I believe that the sum of the lengths of diagonals is maximum if $$\theta_1 = \theta_2 = \dots = \theta_n,$$ which is possible only in a regular polygon.
Edit: same question posted in mathoverflow, https://mathoverflow.net/questions/398248/sum-of-the-lengths-of-all-diagonals-in-a-regular-polygon