There are $\binom {n} {3} = \dfrac {n(n-1)(n-2)} {6}$ conspiring triplets, so (at least) one student $S$ is conspired against by (at least) $\dfrac {(n-1)(n-2)} {6}$ triplets. Since $k$ students can only combine in $\binom {k} {3} = \dfrac {k(k-1)(k-2)} {6}$ triplets, it means those $k$ conspiring against $S$ must be at least such that $\dfrac {(n-1)(n-2)} {6} \leq \dfrac {k(k-1)(k-2)} {6} < \dfrac {k^3}{6}$, hence $k > \sqrt[3]{(n-1)(n-2)}$.

Anyone ??

@above its almost same as #2.Anyway this is a cakewalk for anyone with the basic knowledge of double counting. Italy TST wrote: A stage course is attended by $n \ge 4$ students. The day before the final exam, each group of three students conspire against another student to throw him/her out of the exam. Prove that there is a student against whom there are at least $\sqrt[3]{(n-1)(n- 2)} $conspirators. We count the number of triplets $T=(A,B,C,D)$ where $(A,B,C)$ conspire against $D$.First we choose $(A,B,C)$,this can be done in $\binom{n}{3}$ ways and by the given hypothesis we get there is atleast one such $D$.Hence $T\geq \binom{n}{3}$.Let the number of conspirators against particular person $i$ be atmost $k$.Then after choose $D$ we can select $(A,B,C)$ in atmost $\binom{k}{3}$ ways so $T<n\cdot \binom{k}{3}$.Therefore \[n\cdot \binom{k}{3}>\binom{n}{3} \implies k^3>k(k-1)(k-2)>(n-1)(n-2) \implies k>\sqrt[3]{(n-1)(n-2)}\].Now if $k<\sqrt[3]{(n-1)(n-2)}$ we clearly get a contradiction so there is a person against whom atleast $\sqrt[3][{(n-1)(n-2)}$ people conspired and we are done.$\blacksquare$ REMARK-If we work upon the cubic we get we see that the bound can be incresed to $k=\tfrac{\sqrt[3]{27n^2+\sqrt{(27n^2-81n+54)^2-108}-81n+54}}{3\sqrt[3]{2}}+\tfrac{\sqrt[3]{2}}{\sqrt[3]{27n^2+\sqrt{(27n^2-81n+54)^2-108}-81n+54}}+1$ but since that would be brutual and impossible without wolfram in the TST so I guess the bound given is ok

@above , Thanks , I also think it's equivalent to the solution in post 2 which uses pigeonhole , since the pigeonhole principle can be easily proved using contradiction , I just wanted to make sure that I didn't make any mistakes .

yousseframzi wrote: @above , Thanks , I also think it's equivalent to the solution in post 2 which uses pigeonhole , since the pigeonhole principle can be easily proved using contradiction , I just wanted to make sure that I didn't make any mistakes . I think your solution makes more use of double counting than pigeonhole

cute and simple Let $Q$ be the number of ordered quadruples $(A,B,C,D)$ where the first three defeat $D$. Now, by the first condition $Q= \binom{n}{3}$. Now assume the contrary. We start with a person first. Then we get that $Q \le n. \binom{\sqrt[3]{(n-1)(n-2)}}{3} <n.\frac{(n-1)(n-2)}{6}$,which is a contradiction. Sadly vacation ends..

Call a group of $3$ people a $\emph{clique}$. There are $n \choose 3$ cliques and $n$ people so by pigeonhole there exists a person $P$ with at least $\frac{{n \choose 3}}{n} = \frac{(n-1)(n-2)}{6}$ cliques against them. Let $S$ be the set of people conspiring against this person $P$. Note that the number of cliques formed by people in $S$ is equal to $|S| \choose 3$ In particular, \[ \frac{(n-1)(n-2)}{6} \le {|S| \choose 3} = \frac{|S| \cdot |S|-1 \cdot |S|-2}{6} < \frac{|S|^3}{6}\]It follows that $|S| > \sqrt[3]{(n-1)(n-2)}$ as desired.

Why (n-1)(n-2)/6 ≤(|S|•|S|-1•|S|-2)/6

The person $P$ has at least $\frac{(n-1)(n-2)}{6}$ cliques against them, and all these cliques come from the set $S$ (by definition). Since $S$ contains ${|S| \choose 3} = \frac{|S| \cdot |S|-1 \cdot |S|-2}{6}$ cliques, we must have $\frac{(n-1)(n-2)}{6} \le \frac{|S| \cdot |S|-1 \cdot |S|-2}{6}$

Consider the quadruplets \( (CO, CO, CO, CA) \), where \( CO \) means conspirators, and \( CA \) means conspirated against. Fixing \( (CO, CO, CO) \), there would be \( \binom{n}{3} \) ways. Let the person with the maximum conspirators have \( m \) conspirators, then there are at most \( n \cdot \binom{m}{3} \) ways. So, \( n \cdot \binom{m}{3} \geq \binom{n}{3} \implies m(m - 1)(m - 2) \geq (n - 1)(n - 2) \implies m^3 > (n - 1)(n - 2) \), which gives the result.