Prove that the maximizing point configuration on the unit circle for a Vandermonde like functional is a picket fence For $\lambda_i \in S^1 \subset \mathbb{C}$, consider the functional $H(\{\lambda_1, \ldots, \lambda_n\}):= \sum_{j < k} | \lambda_j - \lambda_k | $. I want to show that $H$ is globally maximized by some picket fence configuration, that is, $\lambda_j = e^{2 \pi i (\alpha + j/n)}$, for some $\alpha \in \mathbb{R}$. 
Let $\lambda_j = e^{i \theta_j}$, with $0 \le \theta_1 \le \ldots \le \theta_n < 2 \pi$. Using the identity $| e^{i \theta_1} - e^{i \theta_2}| = | 2 \sin \frac{\theta_1 - \theta_2}{2}|$ and the fact that $0 \le \theta_k - \theta_j < 2 \pi$, I reduce the problem to maximizing over $\theta_j$ satisfying the above constraints, the following function:
$$ \sum_{j < k} \sin \frac{\theta_j - \theta_k}{2}.$$
Assuming $\theta_j$'s are distinct for now, differentiating gives me
$$ \sum_{\ell < k} \cos \frac{\theta_\ell - \theta_k}{2} - \sum_{j < \ell} \cos \frac{\theta_j - \theta_\ell}{2} = 0, \qquad \forall \ell=1,\ldots, n.$$
But how do I finish?
 A: Simon Rubinstein solved this for me using Jensen's inequality, like a true probabilist would. 
We want to show that the maximizing value is the one attained by a picket fence on $S^1$. Any $n$ points on $S^1$ can be arranged in a counterclockwise order as $\lambda_1, \ldots, \lambda_n$, with some arbitrary starting point $\lambda_1$. Let $\Lambda_{m,i} = \{ \lambda_i , \lambda_{i+m}, \lambda_{i+2m}, \ldots \}$ and $|\Lambda_{m,i}| = k(m)$.  We can reduce the problem to showing that for any $m < n$, the maximizing tuple $(\lambda_i, \lambda_{i+m}, \lambda_{i+2m}, \ldots, \lambda_{i+k(m)m} )$ for the functional $F_{m,i} = \sum_j |\lambda_{i + jm} - \lambda_{i+(j-1)m} |$ is given by the evenly spaced one, i.e., a sub-picket fence. Notice this functional is different from my original (call that $F$), although they are intimately related: $F$ is the sum of a fixed set of $F_{m,i}$'s. So the sum of maximals of the latter is $\geq$ the maximum of the former, though the converse is not necessarily true. 
To show that $F_{m,i}$ is maximized by picket fence, one way is to invoke Jensen (exercise). Alternatively, however, if $\Lambda_{m,i}$ is not a picket-fence, then there will be two adjacent arcs that are unequal in lengths. That is there are three points $\lambda_{i+(j-1)m}, \lambda_{i+jm}, \lambda_{i+(j+1)m}$ such that $|\lambda_{i+(j-1)m} - \lambda_{i+jm}| \neq |\lambda_{i+jm} - \lambda_{i + (j+1)m} |$. Then by moving $\lambda_{i+jm}$ to the middle of $\lambda_{i+(j-1)m}$ and $\lambda_{i+(j+1)m}$ increases the $F_{i,m}$. Thus the only local maximum of $F_{i,m}$ is the picket fence.
