This solution, by Titu Zvonaru, is given in the February 2008 issue of *Crux Mathematicorum*:

Suppose that the most unfortunate player beat 3 others. Label the players so that this player is *P*_{1} and that the players he beat follow him immediately in the lineup, here as *P*_{2}, *P*_{3}, and *P*_{4}. Since the minimum number of wins is 3, there must be some player *P*_{k} such that player *P*_{2} beat player *P*_{k} and *k* is greater than 4 (otherwise player *P*_{2} could have beaten only players *P*_{3} and *P*_{4}, which would give *him* the minimum number of wins, contrary to our supposition). This means that player *P*_{k} must have beaten player *P*_{1}, and we can declare that *A* = *P*_{1}, *B* = *P*_{2}, and *C* = *P*_{k}.

The same argument holds for any minimum number of wins. (It wouldn’t work if the minimum were zero, but we’ve been told that’s not the case.)

09/02/2024 Reader Robert Filman offers an inductive proof:

The theorem is obviously true for *n* = 1, 2 and 3. (*n*=1 is not a possible situation, because everyone won a game, and with *n*=1, there are no games; with *n*=2, there’s only one game, so the loser of that game had no one to beat. With *n*=3, the conditions require *A*>*B*, *B*>*C*, and *C*>*A*.)

Assume hypothesis is true for 1, 2, 3, 4, … *n* – 1 players, and prove it true for *n* players.

Pick any player, *A*. Divide the players into the ones that beat *A* (set *W*) and the ones that lost to *A* (set *L*). *L* is not empty, because *A* beat someone.

If any player *x* in *L* beat any player *y* in *W*, we have a triangle <*a*, *x*, *y*>

If not, consider the set *L*. No one in *L* beat anyone outside of *L*. Every player in *L*’s win came against another player in *L*. But the size of *L* is at most *n* – 1. So by induction, the theorem is true for *L*.

Hence, the theorem is always true.

(Thanks, Robert.)