Problem:
Find all integers n≥3 such that among any n positive real numbers a1,a2,…,an with
max(a1,a2,…,an)≤n⋅min(a1,a2,…,an)
there exist three that are the side lengths of an acute triangle.
Solution:
First we prove that any n≥13 is a solution of the problem. Suppose that a1,a2,…,an satisfy max(a1,a2,…,an)≤n⋅min(a1,a2,…,an), and that we cannot find three that are the side-lengths of an acute triangle. We may assume that a1≤a2≤…≤an. Then ai+22≥ai2+ai+12 for all i≤n−2. Let (Fn) be the Fibonacci sequence, with F1=F2=1 and Fn+1=Fn+Fn−1. It is easy to check that Fn<n2 for n≤11,F12=122 and Fn>n2 for n>12 (the last inequality follows by an immediate induction, while the first one can be checked by hand). The inequality ai+22≥ai2+ai+12 and the fact that a1≤a2≤…≤an imply that ai2≥Fi⋅a12 for all i≤n. Hence, if n≥13, we obtain an2>n2⋅a12, contradicting the hypothesis. This shows that any n≥13 is a solution of the problem.
By taking ai=Fi for 1≤i≤n, we have max(a1,a2,…,an)≤n⋅min(a1,a2,…,an), for any n<13, but it is easy to see that no three ai 's can be the side-lengths of an acute triangle. Hence the answer to the problem is: all n≥13.
This problem and solution were suggested by Titu Andreescu.