As far as I understand, to obtain an optimal solution of the Dual of the MST, meaning:
\begin{align} ~\max &~ z (|V|-1) + \sum_{S \subseteq V : |S| \neq \emptyset} (|S|-1) y_{s} \\ \label{DMST2} s.t. &~ z + \sum_{S: (i,j) \in E(S)}^{} y_{s} \leq w_{ij} & \forall (i,j) \in E \\ \label{DMST3} &~y_{s} \leq 0 & \forall S \subseteq V : |S| \neq \emptyset \end{align}
we have to start with the trivial partition and add tight edges to it, one by one.
What I don't undestand is how we obtain those tight edges. Suppose we have the $K_5$ with the edgweights being the sum of the corresponding vertices (vertices being $1,2,3,4,5$). How can I see that $(1,2)$ is a tight edge?