Prove that the maximizing point configuration on the unit circle for a Vandermonde like functional is a picket fence This is probably too hard for math.stackexchange, so I migrated it here.
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: Use Hadamard's inequality for the determinant 
$$
   | \det (\mathbf{a}_1, \mathbf{a}_2, \ldots , \mathbf{a}_n ) |
 \leq \| \mathbf{a}_1 \| \| \mathbf{a}_2 \| \cdots \| \mathbf{a}_n \| ,
$$
with equality if and only if $\mathbf{a}_1, \mathbf{a}_2, \ldots , \mathbf{a}_n $ 
forms an orthogonal system. 
Then 
$$
 |H(\lambda_1, \ldots ,\lambda_n )| 
 = abs \det 
\left( \begin{array}{cccc}
1 & \lambda_1 & \cdots & \lambda_1^{n-1} \\
 1 & \lambda_2 & \cdots & \lambda_2^{n-1} \\
\vdots & \vdots & \ddots & \vdots \\
1 & \lambda_n & \cdots & \lambda_n^{n-1}
\end{array} 
\right)
 \leq n^{\frac{n}{2}}
$$
with equality if and only if 
\begin{align*}
& \lambda_1+ \lambda_2 + \cdots + \lambda_n = 0  \\
& \lambda_1^2+ \lambda_2^2 + \cdots + \lambda_n^2 = 0  \\
& \vdots   \\
& \lambda_1^{n-1} + \lambda_2^{n-1} + \cdots + \lambda_n^{n-1} = 0 .
\end{align*}
Equivalently elementary symmetric polynomials up to order $n-1$ vanish, i.e.,
\begin{align*}
\sum_{k=1}^n \lambda_k = 0, \quad
\sum_{1\leq k_1< k_2 \leq n} \lambda_{k_1} \lambda_{k_2} = 0, \ldots , \\
\lambda_2 \lambda_3 \cdots \lambda_n + \lambda_1 \lambda_3 
\cdots \lambda_n + \cdots + \lambda_1 \lambda_2 \cdots \lambda_{n-1} = 0  .
\end{align*}
This means $\lambda_1, \lambda_2 , \ldots ,  \lambda_n $ forms 
a picket fence in your notation.
