# Show $\sum_{i \in V(G)}(out(i)-in(i))^3 \geq 0$ for a complete directed graph G with more than 3 vertices

Given a complete graph $$G$$ on $$n$$ vertices and we assign direction to edges. Suppose that for any set of four vertices, it is not true that there is a vertex among the four with only in-edges and the edges between the rest of the three vertices form a directed triangle. Show that $$S(G):=\sum_{i \in V(G)}(out(i)-in(i))^3 \geq 0$$

where $$out(v)$$ and $$in(v)$$ are the out degree and in degree of $$v$$ respectively.

Solution attempt:

if $$|G|=4$$, then the claim is true.

We look at case for $$|G|>4$$. For node $$i \neq j$$, let $$\epsilon_{i,j}=1$$ if there is an edge from $$i$$ to $$j$$ and $$\epsilon_{i,j}=-1$$ if there is an edge from $$j$$ to $$i$$. So $$(out(i)-in(i))^3 =(\sum_{j\neq i}\epsilon_{i,j})^3=\sum_{j_1,j_2,j_3 \neq i}\epsilon_{i,j_1}\epsilon_{i,j_2}\epsilon_{i,j_3}$$.

So $$S(G) = \sum_{i \notin \{j_1,j_2,j_3\}}\epsilon_{i,j_1}\epsilon_{i,j_2}\epsilon_{i,j_3}$$