We induct on $n$ (the size of the tournament), the base case $n=3$ being trivial. Consider a tournament with $k+1$ players. Let $A$ be the player who has won the maximum number of games. If $A$ has defeated all the other $k$ players, we're done by the inductive hypothesis on a tournament of size $k$. Otherwise, suppose there exists a player $C$ who defeats $A$. If there existed no player $B$ such that $A$ defeated $B$ and $B$ defeated $C$, then $C$ would have defeated more players than $A$, which is a contradiction. $\blacksquare$

Let the players be $P_1, P_2, \cdots , P_n$. For all $i$, let $f(i)$ be the number of players $P(i)$ has defeated. Call a triple $i$-bad if it is not a cycle, and more specifically, if $P_i$ is the player in that triple who has defeated the other two players. Note that there are exactly $\dbinom{f(i)}{2}$ $i$-bad triples, so the total number of bad triples is $\displaystyle \sum_{i=1}^{n} \dbinom{f(i)}{2}$. Since the number of cycles is equal to $\dbinom{n}{3}$ minus the total number of bad triples, we need to minimize $\displaystyle \sum_{i=1}^{n} \dbinom{f(i)}{2}$. Note that $\displaystyle \sum_{i=1}^{n} f(i) = \dbinom{n}{2}$ since both sides are equal to the total number of games. Now, by Cauchy, \[\begin{align*} \dfrac{1}{2} \displaystyle \sum_{i=1}^{n} f(i)^2 - f(i) & \geq \dfrac{1}{2} \left( \displaystyle \dbinom{n}{2}^2 \cdot \dfrac{1}{n} - \dbinom{n}{2}\right) \\ & = \dfrac{n(n-1)(n-3)}{24}.\end{align*}}\] Hence, the total number of cycles is at least $\dbinom{n}{3}- \dfrac{n(n-1)(n-3)}{24} = \dfrac{(n-1)n(n+1)}{24}.$ Equality holds when all players have won the same number of games (note that this subsequently implies $n$ to be odd, as $\displaystyle \sum_{i=1}^{n} f(i) = nf(1) = \dbinom{n}{2} \implies n \mid \dbinom{n}{2}$, which implies $24 \mid n(n-1)(n+1)$).