C2. The edges of $K_{2017}$ are each labelled with 1,2 , or 3 such that any triangle has sum of labels at least 5 . Determine the minimum possible average of all $\binom{2017}{2}$ labels.
(Michael Ma)
In general, the answer for $2 m+1$ is $2-\frac{1}{2 m+1}$.
We prove the lower bound by induction on $m$ : assume some edge $v w$ is labeled 1 . Then we delete it, noting that edges touching $v$ and $w$ contribute a sum of at least $4 \cdot(2 m-1)=8 m-4$. Thus by induction hypothesis the total is at least
$$
\binom{2 m-1}{2}\left(2-\frac{1}{2 m-1}\right)+(8 m-4)+1=\binom{2 m+1}{2}\left(2-\frac{1}{2 m+1}\right)
$$
as desired.
Interestingly, there are (at least) two equality cases. One is to have all edges be 2 except for $m$ disjoint edges, which have weight 1 . Another is to split the vertex set into two sets $A \cup B$ with $|A|=m$ and $|B|=m+1$, then weight all edges in $A \times B$ with 1 and the remaining edges with 3 .

Remark. In fact, given any equality case on $c$ vertices, one can generate one on $c+2$ vertices by two vertices $u$ and $v$, connected to the previous $c$ vertices with weight 2 , and then equipping $u v$ with weight 1 .