In a finite directed complete graph $G ( V, E )$, if all edges have weight either $1$ or $2$, how to show that weights of edges of $G$ satisfies "Triangular Inequality"?
Edited
Where triangular inequality is weight of any edge is less than or equal to the sum of weights of any other two edges (ALWAYS)
given an edge, it has weight 1 or 2, that is always less or equal to the sum of the weights of two edges.
This means that, if we define $d(u,v)$ as the minimum distance between the nodes $u$ and $v$, we have $1\le d(u,v)\le 2$, and so $$d(u,v)\le 2=1+1\le d(u,w)+d(w,v)$$ since every pair of nodes is connected by an edge.