WakeUp wrote: Determine all positive integers $n\ge 3$ such that the inequality \[a_1a_2+a_2a_3+\ldots a_{n-1}a_n\le 0\] holds for all real numbers $a_1,a_2,\ldots , a_n$ which satisfy $a_1+a_2+\ldots +a_n=0$. For $n=3$, we have $a_1a_2+a_2a_3=-a_2^2\le 0$. For $n>3$ it is false, take $(2-n,0,1,...,1)$.

Actually, the inequality is still true for $n=4$ since we have \[a_1a_2+a_2a_3+a_3a_4+a_4a_1=(a_1+a_3)(a_2+a_4)=-(a_1+a_3)^2 \le 0.\]Indeed, your counterexample does not work for any single $n$ since it produces the value $-1$ which is fine. However, one can take $(1,1,0,-2,0,0,\dots,0)$ as an counterexample for any $n \ge 5$.

WakeUp wrote: Determine all positive integers $n\ge 3$ such that the inequality \[a_1a_2+a_2a_3+\ldots a_{n-1}a_n\le 0\]holds for all real numbers $a_1,a_2,\ldots , a_n$ which satisfy $a_1+a_2+\ldots +a_n=0$. The problem was: Determine all positive integers $n\ge 3$ such that the inequality \[a_1a_2+a_2a_3+\cdots+a_{n-1}a_n+a_na_1\le 0\]holds for all real numbers $a_1,a_2,\cdots , a_n$ which satisfy $a_1+a_2+\cdots +a_n=0$.

Actually, you still got it wrong, since the last term should be $a_na_1$. But yes, thanks for this point! My solution is the correct one for the actual contest problem while the solution in #2 is the correct one for the problem in #1.

WakeUp wrote: Determine all positive integers $n\ge 3$ such that the inequality \[a_1a_2+a_2a_3+\ldots a_{n-1}a_n\le 0\]holds for all real numbers $a_1,a_2,\ldots , a_n$ which satisfy $a_1+a_2+\ldots +a_n=0$. In the Baltic Way problem collection is the wrong problem. (The one given in #1). Can someone fix it.