Sum of the lengths of all diagonals in a regular polygon 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.
 A: I am going to assume that all the points lie on the complex unit circle. The points are denoted by $e^{i x_k}$ for $k=1$ to $n$ in the counter-clockwise direction. All indices are mod $n$ for simplicity. The sum of distances that we are looking to minimize is a sum in the following form:
$$\sum_{r,t}\sqrt{2-2\cos(x_r-x_t)}$$
This sum can be broken down to sums of the following form where $m$ ranges from $1$ to $\lfloor n/2 \rfloor$:
$$s_m = \sum_{r}\sqrt{2-2\cos(x_{r+m}-x_{r})}$$
I claim that each $s_m$ gets maximized when the polygon is regular. This is a constrained optimization problem, constrained by $\sum x_{r+m}-x_r=2m\pi$. The proof is very simple, by Jensen's inequality: the function $f(x)=\sqrt{1-\cos x}$ is concave as its second derivative is $\frac{-1}4\sqrt{1-\cos x}$. Jensens's inequality gives $s_m \le n f(2m\pi/n)$, with equality when $x_{r+m}-x_r=2m\pi/n$ for all $r$. This happens when the polygon is regular (there might be other configurations that this might happen for example in the quadrilateral case the sum of diagonals of rectangle is equal to a square).
