Happy ants never leave compact domain? I am curious if the following seemingly simple question has an easy answer?
Consider an ant population of $N$ ants that lives in $\mathbb R^2$. Each ant can be labeled by some coordinate $x\in \mathbb R^2.$
Ants like to be close to their peers but also not too close. The optimal distance between the center of two ants is $5^{2/3}$. So given two ants $x_i,x_j \in \mathbb R^2$. Their happiness $H$ is $$H(x_i,x_j):=\operatorname{max}\{-\vert \vert x_i-x_j \vert^{3/2} -5 \vert,-10\}.$$
Distances $\vert x_i-x_j\vert \le 1$ are not allowed since ants do not like to get too close.
The purpose of the maximum, in the definition of $H$ above, reflects that there is no attraction between two ants anymore at a certain distance.
Now consider the total happiness $H_N:=\sum_{i<j} H(x_i,x_j).$ The question is the following:
Let $x=(x_1,..,x_N)$ be any global minimizer of the happiness $H_N.$ Can we find a radius $r$ such that all ants of a global maximiser of the happiness are within a distance $r\sqrt{N}$ from one another?
The scaling $r\sqrt{N}$ is due to the fact that order $N$ lattice particles would fit into a ball of radius $r\sqrt{N}.$
The question sounds almost like a school problem, and may admit a very simple solution, if one finds the right way of arguing here, but it seems just hard to exclude the huge amount of bizarre configurations.
 A: I am going to describe a partial solution / proposal to obtain a solution, which I think is interesting enough to post even though it doesn't fully answer the question.

First, I wish to redefine $H(x_i,x_j)$ as $\max \{ 10 - | | x_i - x_j|^{3/2}-5|,0\}$, i.e. add $10$ to make it nonnegative.
Having done this, we obtain $\max H_{N_1+N_2} \geq \max H_{N_1} + \max H_{N_2}$, so $\lim_{N \to \infty} \frac{ \max H_N}{ N} $ exists and is either finite or $+\infty$. There is some limit to the happiness $\sum_j H(x_i,x_j)$ of an individual ant based on the number of discs of radius $1/2$ we can pack into a disc of radius $15^{2/3}$, so it is not $+\infty$ and thus is finite. Call this $\lambda$.

My approoach rests on the following conjecture:

There exists a constant $c>0$ such that, for any configuration of $N$ ants, such that the graph with vertices ants and edges pairs of ants separated by a distance of at most $15^{2/3}$ is connected, the total happiness is at most $\lambda N - c R$ where $R$ is the minimal radius of a disc enclosing all the ants.

The motivation for this conjecture is that, whatever configuration gives the maximum $\lambda$ happiness per ant, the boundary of the ant colony will form a flaw in that configuration, leading to a loss from the maximal $\lambda N$ happiness proportional to the size of the boundary. Under this mild connectedness hypothesis, $R$ is bounded by the size of the boundary, explaining the lost happiness proportional with $R$.
Combined with the statement that, for all $N$, there exists a configuration of happiness at least $\lambda N - O (\sqrt{N})$, this implies an upper bound as you desire.
However, I am not sure how to prove the conjecture without a more precise understanding of the optimal configuration - it is possible to imagine, say, packing ants in some highly efficient way which can't be continued past a certain curve, leading to very happy configurations with large boundary.

I will now prove the existence statement.
Take a configuration of a very large number of ants that attains an average happiness of $\lambda-\epsilon$. Let $r$ be minimal such that a ball of radius $r$ can pack at most $n$ ants. Consider a random disc of radius $r$ in this large configuration. The expected number of ants in this disc is $ d \pi r^2$ where $d$ is the density. The expected total happiness of the ants in the disc is $(\lambda-\epsilon) d \pi r^2$. The expected number of ants in the disc within a distance $15^{2/3}$ of the boundary is $ O ( d \pi r)$ and so the expected loss to the happiness of the ants in the disc from removing all the ants outside the disc is $O( d \pi r)$. So the expected total happiness of the ants in the disc, once the outside ants are removed, is $$ \geq (\lambda - \epsilon ) d \pi r^2 - O( d \pi r).$$
So there must exist some configuration of $\leq n$ ants whose average happiness is at least
$$ \frac{ (\lambda - \epsilon ) d \pi r^2   - O( d\pi r)}{  d \pi r^2} = \lambda - \epsilon - O \left( \frac{1}{r} \right)$$
We have $r \approx \sqrt{n}$ so this is $$\lambda - \epsilon - O \left(\frac{1}{\sqrt{n} }\right).$$ Taking $\epsilon$ to $0$, we see that there is a configuration of $m \leq n$ ants with happiness $\geq m \lambda - O\left( \frac{m}{\sqrt{n}}\right)$.
I claim that there is a configuration of exactly $N$ ants with happiness $\lambda N - O ( \sqrt{N})$. To see this, take $n_1= N$ and find a configuration of $m_1$ ants with average happiness $\geq \lambda - O( \frac{1}{\sqrt{n_1}})$, then repeat it $ \lfloor \frac{n_1}{m_1} \rfloor$ times, leaving room for $n_2 = n_1 - m_1 \lfloor \frac{n_1}{m_1} \rfloor$ ant. Then apply the previous existence result again to find a configuration of at most $n_2$ ants with average happiness $\geq \lambda - O ( \frac{1}{ \sqrt{n_2} })$ , and iterate. The total happiness is $$\geq \lambda N - \sum_i  \left\lfloor \frac{n_i}{m_i} \right\rfloor O \left( \frac{m_i}{ \sqrt{n_i}}\right) \geq \lambda N - \sum_i O \left(\sqrt{n_i} \right) = \lambda N - O (\sqrt{N})$$ since $n_{i+1} < n_i/2$ so $n_i < N/ 2^{i-1}$.
