Does the electrostatic potential have a local maximum on the sphere? Let
$$M=\{(x_1,x_2,x_3,x_4) \in  \mathbb{S}^2 \times \mathbb{S}^2 \times \mathbb{S}^2 \times \mathbb{S}^2 \, |\,\, \text{ all the } x_i \, \text{ are distinct}\} $$ $M$ is an open subset of $( \mathbb{S}^2)^4$.
Let $E:M \to \mathbb{R}$ be defined by
$$E(x_1,x_2,x_3,x_4)=\sum_{i < j}\frac{1}{\| x_i - x_j \|},$$
where $\| x_i - x_j \|$ denotes the Euclidean distance in $\mathbb{R}^3$.
Question: Does $E$ have a point of local maximum?
Clearly $E$ doesn't have a global maximum, since it's unbounded. Intuitively, given any point $p \in M$, we can move some of the $x_i$ closer to each other, thus increasing the energy. However, when we move say $x_2$ closer to $x_1$, we might be pushing it further away from $x_3$ or $x_4$, so there are "competing" changes in the contribution of each $\| x_i - x_j \|^{-1}$ term.
Is there an easy way to see that one can always choose a direction where $E$ strictly increases?
 A: Nope.
Given any two points $p,q \in S^2$ separated by an angle $\theta \in (0,\pi]$, let $c = \cos\theta$ and $s = \sin\theta$. We have
$$\frac{1}{|p-q|} = \frac{1}{\sqrt{2(1-c)}}$$
Fix $q$ and perturb $p$ by a small amount to $p' = p + \delta p\in S^2$ so that $c \to c + \delta c$, we have
$$\delta\left[\frac{1}{|p - q|}\right] \stackrel{def}{=} \frac{1}{|p' -q|}
- \frac{1}{|p-q|} = \frac1{\sqrt{2}}\left(\frac{\delta c}{2\sqrt{1-c}^3} + \frac{3(\delta c)^2}{8\sqrt{1-c}^5} \right) + O((\delta c)^3)\tag{*1}
$$
Let $a \in S^2$ so that $a \perp p$ and $q = c p + s a$, Let $b = p \times a$.
For any $0 < \epsilon \ll \theta$, consider perturbations of $p$ to a point $p'$ at angle $\epsilon$ from it.
ie. pick a $\phi$ from $[0,2\pi)$ and set
$$p' = P(\epsilon,\phi) \stackrel{def}{=} \cos\epsilon p + \sin\epsilon (a \cos\phi + b \sin\phi)$$
The corresponding $\delta c$ equals to $c(\cos\epsilon - 1) + s\sin\epsilon\cos\phi$.
Substitute this into $(*1)$ and only keeping terms up to $\epsilon^2$, we obtain
$$\delta\left[\frac{1}{|p-q|}\right]_{p'=P(\epsilon,\phi)}
= \frac1{\sqrt{2}}\left[\frac{2s\epsilon\cos\phi - c\epsilon^2}{4\sqrt{1-c}^3}
+ \frac{3(s\epsilon\cos\phi)^2}{8\sqrt{1-c}^5}
\right] + O(\epsilon^3)$$
Taking angular average over $\phi$, we obtain
$$\begin{align} \frac{1}{2\pi} \int_0^{2\pi} \delta\left[\frac{1}{|p-q|}\right]_{p'=P_(\epsilon,\phi)} d\phi
& = \frac{\epsilon^2}{\sqrt{2}}\left[-\frac{c}{4\sqrt{1-c}^3} + \frac{3s^2}{16\sqrt{1-c}^5} \right] + O(\epsilon^3)\\
&=  \frac{\epsilon^2}{\sqrt{2}}\frac{3-c}{16\sqrt{1-c}^3} + O(\epsilon^3)
\end{align}
$$
Since $\displaystyle\;\frac{3-c}{16\sqrt{1-c}^3} > 0\;$, we find for all sufficient small $\epsilon$,
$$\frac{1}{2\pi}\int_0^{2\pi} \frac{d\phi}{|P(\epsilon,\phi) - q|} > \frac{1}{|p-q|}\tag{*2}$$
For any $4$ points $p_1,\ldots,p_4$, replace $p$ by $p_1$ and $q$ by $p_2,p_3,p_4$ and sum over the pairs.
$(*2)$ tell us for all sufficient small $\epsilon$, the electrostatic potential energy
$$E(p_1,p_2,p_3,p_4) \stackrel{def}{=} \sum_{i<j} \frac{1}{|p_i - p_j|}$$
satisfies
$$\frac{1}{2\pi} \int_0^{2\pi}E( P(\epsilon,\phi),p_2, p_3, p_4 ) > E(p_1,p_2,p_3,p_4)$$
For such $\epsilon$, we can always find a point $p'$ at an angle from $p_1$ with $$E(p_1',p_2,p_3,p_4) > E(p_1,p_2,p_3,p_4)$$
This means $p_1,p_2,p_3,p_4$ cannot be a local maximum for $E(\cdots)$.
Since this is true for any $4$ points on $S^2$, $E(\cdots)$ doesn't have any local maximum.
