maximize $\sum_{A\subseteq [q], A\neq \emptyset} \alpha_A \log(|A|)$ with nonlinear constraints Let $[q]:=\{1,2,3, \ldots,q\}$, where $q$ is a positive integer. Consider a vector $\underline{\alpha}=(\alpha_A)_{A\subseteq [q], A\neq \emptyset}$, where each $\alpha_A \in \mathbb{R}$. Note that such a vector $\underline{\alpha}$ has $2^q-1$ entries.
Given such an $\underline{\alpha}$, we define the following:
$$ OBJ(\underline{\alpha}):=\sum_{A\neq \emptyset} \alpha_A \log(|A|), \quad v(\underline{\alpha})=\sum_{A\neq \emptyset} \alpha_A, \quad E(\underline{\alpha})=\sum_{\{ \{A,B\}: A,B \subseteq [q], A\cap B=\emptyset \}} \alpha_A \alpha_B, $$
where the sum in the definition of $E(\underline{\alpha})$ is taken over all unordered pairs of disjoint subsets of $[q]$.
Define $FEAS(1/4)=\{\underline{\alpha}: \alpha_A \geq 0 \text{ for all nonempty A },\,  v(\underline{\alpha})=1, \, E(\underline{\alpha})\geq 1/4 \}$.
I believe that the following is true:
$$OPT(1/4):= \max_{\underline{\alpha} \in FEAS(1/4) } OBJ(\underline{\alpha})=\frac{\log(\lfloor q/2 \rfloor \cdot \lceil q/2 \rceil)}{2}.   $$
I know that it is true when $q$ is even (it was proven), but I want to show that it also holds for odd $q$. I have verified that this is true for $q=3,5,7,9$ using SageMath, but would like to prove it by hand for general odd $q$.
I think that this problem can be solved using the language of probability. It is natural to do so given that the sum $v(\underline{\alpha})=\sum \alpha_A=1$ in the set $FEAS(1/4)$.
Consider a random variable $X$ on $2^{q}\setminus \{\emptyset\}$ (the power set of $[q]$, excluding the empty set) such that $\mathbb{P}[X=A]=\alpha_A$, where $A \subseteq [q]$.
Note that $2\cdot E(\underline{\alpha})$ can be interpreted as the probability that from the sets in $2^{[q]}\setminus \{\emptyset\}$ one selects two disjoint sets $A$ and $B$. Since by definition of $FEAS(1/4)$, $2\cdot E(\underline{\alpha}) \geq 1/2$, we see that we are more likely to select two disjoint sets rather than two sets which have a nonempty intersection.
One can verify that the set $FEAS(1/4)$ is compact, so there exists some $\underline{\alpha}^*\in FEAS(1/4)$ for which $F(\underline{\alpha}^*)=\max_{\underline{\alpha} \in FEAS(1/4) } F(\underline{\alpha})$. I conjecture that this vector $\underline{\alpha}^*$ has exactly 2 nonzero entries $\alpha^*_{A_1}=1/2$ and $\alpha^*_{A_2}=1/2$, where $|A_1|=\lfloor q/2 \rfloor$, $|A_2|=\lceil q/2 \rceil$, $A_1\cap A_2 =\emptyset$, and $A_1\cup A_2=[q]$. That is, the sets $A_1$ and $A_2$ form a partition of the first $q$ positive integers and their sizes are as equal as possible.
If it is difficult to prove this for general odd $q$, how would I prove it for, say, $q=5$? I would like to avoid using Lagrange multipliers with so many variables.
Let $\gamma \in \mathbb{R}$ such that $0\leq \gamma \leq \frac{q-1}{2q}$. Then
$$OPT(\gamma)=\{\underline{\alpha}: \alpha_A \geq 0 \text{ for all nonempty A },\,  v(\underline{\alpha})=1, \, E(\underline{\alpha})\geq \gamma \}.$$
In the paper for which I provided the link above, Serguei Norine shows that $OPT(\gamma)\leq \log(q(1-2\gamma))$   , with equality holding if and only if $\gamma=\frac{r-1}{2r}$ for some positive integer $r$ dividing $q$. When $r=2$ we see that $\frac{r-1}{2r}=1/4$, so $OPT(1/4)=\log(q/2)$ if and only if $q$ is divisible by two.
 A: This is not an answer but some thoughts.
It seems difficult to me to verify that OPT even for $q=5.$ What you have is, naively, a nonconvex optimization problem over $2^q-1$ variables. So without some extra insight it seems hopelessly difficult computationally.
Here is an idea for a stronger conjecture that can be verified as a nonconvex optimization in $q^2$ dimensions, more specifically testing copositivity of a $q^2\times q^2$ matrix. That's still very difficult computationally, but slightly better in theory.
Let $$\hat E(\underline\alpha)=\sum_{A,B\in 2^{[q]}\setminus\{\emptyset\}}\alpha_A\alpha_B\frac{|A\cap B|}{\min(|A|,|B|)}.$$
$\hat E(\underline\alpha)$ is a lower bound on $1-2E(\underline\alpha).$ Optimistically we might have
$$OBJ(\underline\alpha)\leq mq\hat E(\underline\alpha)+c\qquad(?)$$
where $m,c$ are the unique numbers satisfying $mx+c=\log x$ for $x\in \{\tfrac{q-1}2,\tfrac{q+1}2\}.$ This would imply your OPT by plugging in $\hat E(\underline\alpha)=\tfrac12.$
Homogenizing, we want
$$OBJ(\underline\alpha)v(\underline\alpha)\leq mq\hat E(\underline\alpha)+cv(\underline\alpha)^2\qquad(?)$$
for all vectors $\alpha$ with non-negative entries.
The functions $OBJ,v,\hat E$ are homogeneous linear, linear, and quadratic functions respectively of the $q^2$ numbers
$$p_{i,j}(\underline\alpha)=\sum_{\substack{|A|=i\\A\ni j}}\alpha_A/|A|$$
In particular
$$\hat E(\underline\alpha)=\sum_{i,j,k=1}^q\max(i,k)p_{i,j}(\underline\alpha)p_{k,j}(\underline\alpha).$$
So this reduces to a slightly easier problem computationally. 
