Increasing sum of edge weights on a vertex and edge weighted graph Let $[n]=\{1, 2, \ldots, n\}$, where $n \geq 3$ is an integer. Let $\mathcal{P}(n)$ be the power set of $[n]$. We define a graph $G$ as follows:
The vertex set of $G$ is $V(G)=\mathcal{P}(n)\setminus \{\emptyset\}$, and its edge set $E(G)=\{AB: A\cap B=\emptyset \}$. 
Notice that $G$ will always have an isolated vertex, namely the vertex $[n]$, because it cannot be disjoint to any of the other sets in $V(G)$.
Now suppose that to each vertex $A$ we associate with it a $\textbf{nonnegative}$ vertex weight $\omega_A$, where $\displaystyle \sum_{A \in V(G)} \omega_A=1$ and $\omega_A$ is a real number. 
To each edge $e=AB$ $\,$ in $\,$ $G$, we give the weight $\omega_A\omega_B$, that is, the product of its vertex ends. Note that since the $\omega_A$'s are nonnegative, some of the edge weights may be zero.
Consider the sum $\displaystyle \sum_{AB \in E(G)} \omega_A\omega_B$. 
I want to find another set of nonnegative weights $\{\beta_A: A \in V(G)\}$, such that $\sum_{A\in V(G)} \beta_A=1$ and such that the $\beta_A$'s are associated to $V(G)$ and $E(G)$ in the same way as the $\omega_A$'s, but
$$\displaystyle \sum_{AB \in E(G)} \omega_A\omega_B \leq \sum_{AB \in E(G)} \beta_A\beta_B.$$
Edit: Assume that $\beta_A\neq \omega_A$ for each $A \in V(G)$ (otherwise the answer is trivial).
 A: Let $\vec\omega$ denote the vector of values $(\omega_A)$, where we always require each $\omega_A \ge 0$ and $\sum_A \omega_A = 1$.  Define:
$$S(\vec\omega) = \sum_{AB \in E(G)} \omega_A \omega_B$$

Claim: $S(\vec\omega)$ is globally maximized when $\omega_A = 1/n$ for $|A| = 1$ and $\omega_A = 0$ for $|A| > 1$.  

First, suppose $\omega_A > 0$ for some $|A| > 1$.  Pick any $C \subsetneq A$ where $|C| = 1$.  Now consider a different vector $\vec\beta$ where:


*

*$\beta_A = 0, \beta_C = \omega_A + \omega_C$

*$\beta_B = \omega_B$ for any other $B$.
I.e., $\vec\beta$ is the same as $\vec\omega$ except the weight at $A$ has been moved to $C$ (added to any weight at $C$).  Now consider the change $S(\vec\beta) - S(\vec\omega)$.  Any term not involving $A$ nor $C$ has not changed.  Terms involving $A$ has decreased to $0$, and terms involving $C$ has increased in value.
$$
\begin{array}{}
S(\vec\beta) - S(\vec\omega) &= \displaystyle \sum_{P: ~CP \in E(G)} (\beta_C - \omega_C)\omega_P - \sum_{Q: ~AQ \in E(G)} \omega_A \omega_Q \\
&= \displaystyle\sum_{P: ~CP \in E(G)} \omega_A\omega_P - \sum_{Q: ~AQ \in E(G)} \omega_A \omega_Q \\
&= \omega_A \Bigl( \displaystyle\sum_{P: ~CP \in E(G)} \omega_P - \sum_{Q: ~AQ \in E(G)} \omega_Q \Bigr) \\
&\ge 0
\end{array}
$$
where the last inequality comes from the fact that $ A\cap Q = \emptyset \implies C\cap Q = \emptyset $ (because $C \subset A$), so $\{Q: AQ \in E(G)\} \subsetneq \{P: CP \in E(G)\}$.
Because $S(\vec\beta) \ge S(\vec\omega)$, this means when looking for global max, there is no need to consider any $\vec\omega$ where $\omega_A > 0$ for some $|A| > 1$.  Any weight assigned to a non-singleton subset $A$ can be re-assigned to some singleton $C\subset A$, without decreasing $S()$.
Thus from now on we only consider $\vec\omega$ where $\omega_A =0$ for all $|A| > 1$.  Such a $\vec\omega$ is fully specified by $n$ weights, i.e. those assigned to the $n$ singleton subsets.  By a slight abuse of notation, we will denote these weights $\omega_1, \omega_2, \dots, \omega_n$.  Using the fact $\sum_i \omega_i = 1$, we have:
$$S(\vec\omega) = \sum_{i > j} \omega_i \omega_j = \frac12 \sum_i \omega_i (1 - \omega_i) = \frac12 (1 - \sum_i \omega_i^2)$$
If two of the weights were unequal, say $\omega_k > \omega_l$, then redistributing the weights equally as $\beta_k = \beta_l = {\omega_k + \omega_l \over 2}$ (and changing nothing else) would reduce the sum of squares, and therefore $S(\vec\beta) > S(\vec\omega)$.  This can keep happening until all the weights are equal, i.e. $1/n$, at which point global max is achieved.  $~~~~\square$
