# Does $(\sum_{i=1}^n a_i^{1.5})^2 - \sum_{i=1}^n a_i \; \sum_{i=1}^n a_i a_{i+1} > 0$ hold for $n \leq 8$?

Let positive reals $\{a_i\}$, where not all $a_i$ are equal. Does $$f(\{a_i\}) = (\sum_{i=1}^n a_i^{1.5})^2 - \sum_{i=1}^n a_i \; \sum_{i=1}^n a_i a_{i+1} > 0$$ hold for $n \le 8$? It is understood that $a_{n+1} = a_1$.

The restriction to $n \le 8$ comes from a known counterexample for $n = 9$ given by Martin R. in this post: $a_i = (40, 37, 40, 50, 60, 65, 65, 56, 47)$. In there, he also gave counterexamples for higher $n$.

Further, there is a proof for $n=3$ in here and Michael Rozenberg has added comments in here that for $n=4$ a solution can be found by $AM-GM$ and in here that for $n=5$, Buffalo Way can produce a solution.

From the counterexamples and from numerical experiments, it appears that the solution will have to do with oscillations, so a Fourier series approach might be helpful.

• Hello, maybe it could help vixra.org/pdf/1008.0030v1.pdf . See at theorem 2 . – max8128 Jan 25 '18 at 13:55
• @max8128 Thanks - I'll check it. – Andreas Jan 25 '18 at 14:18