Number of directed triangles in a tournament?

Question

Observing that each transitive triangle contains exactly one directed 2-path and that each directed triangle contains exactly three directed 2-paths, deduce that the number of directed triangles in $T$ is at most $\frac{1}{4}\binom{n+1}{3}$. When does equality hold?

Answer 1

First, choose a vertex $v$ and let $k$ be the number of ingoing edges on $v$. Then the number of $2$-paths with $v$ as the middle vertex is $k(n-1-k)$. By taking the derivative, you can calculate that this quantity attains it maximum when $k=\frac{n-1}{2}$, and therefore the number of $2$-paths with $v$ as a middle vertex is $\leq\frac{(n-1)^{2}}{4}$.

Hence, for a given tournament $T$ on $n$ vertices, the total number of $2$-paths is $\leq\frac{n(n-1)^{2}}{4}$.

Now we will count the number of $2$-paths by adding those found in the directed triangles and those from the transitive triangles. Let $m_{d}$ be the number of directed triangles. Then by the above, \begin{equation} \left(\binom{n}{3}-m_{d}\right)+3m_{d}\leq\frac{n(n-1)^{2}}{4}, \end{equation} which simplifying gives $m_{d}\leq\frac{1}{4}\binom{n+1}{3}$.

As for the equality, every vertex would need to have $\frac{n-1}{2}$ ingoing and outgoing edges.