we count the number of triangles in the directed graphs so we get that: $\binom{2n+1}{3}=T+S$ where $T$ is the desired triangles and $S$ are the other triangles... now every triangle in $S$ is like this:$A$ has won from $B,C$ so we can count $S$ acording to vertex $A$ of it... so $S=\sum_{i=1}^{2n+1}\binom{d_{i}}{2}$ where $d_{i}$ is the number of games that vertex $i$ has lost... now we know that $\sum_{i=1}^{2n+1}d_{i}=\binom{2n+1}{2}=n(2n+1)$so we get that $S=\frac{\Sigma d_{i}(d_{i}-1)}{2}=\frac{\Sigma d_{i}^{2}-\Sigma d_{i}}{2}=\frac{\Sigma d_{i}^{2}-n(2n+1)}{2}$ so $S$ is minimum iff $\Sigma d_{i}^{2}$ is minimum,but we know that $\Sigma d_{i}^{2}=\frac{(d_{1}+...+d_{2n+1})^{2}}{2n+1}=n^{2}(2n+1)$ so we get that: $T=\binom{2n+1}{3}-S\leq \binom{2n+1}{3}-\frac{n^{2}(2n+1)+n(2n+1)}{2}=\frac{6n^{3}+n^{2}-n}{6}$