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.
Remarks: As @Michael Rozenberg pointed out, the case $n=3, 4$ can be proved nicely by the power mean inequality, and the case $n=5$ can be proved by the Buffalo Way (BW). Actually, the case $n=6$ can be also proved by BW. I believe the cases $n=7, 8$ can be proved by BW. I leave them for the author of the OP.
Case $n=3$: Since $a_1^2a_2^2 + a_2^2a_3^2 + a_3^2 a_1^2 \le \frac{(a_1^2 + a_2^2 + a_3^2)^2}{3}$, it suffices to prove that $(a_1^3 + a_2^3 + a_3^3)^2 \ge \frac{(a_1^2 + a_2^2 + a_3^2)^3}{3}$ which is true by the power mean inequality.
Case $n=4$: Since $a_1^2a_2^2 + a_2^2a_3^2 + a_3^2a_4^2 + a_4^2a_1^2 = (a_1^2 + a_3^2)(a_2^2 + a_4^2) \le \frac{(a_1^2 + a_2^2 + a_3^2 + a_4^2)^2}{4}$, it suffices to prove that $(a_1^3 + a_2^3 + a_3^3 + a_4^3)^2 \ge \frac{(a_1^2 + a_2^2 + a_3^2 + a_4^2)^3}{4}$ which is true by the power mean inequality.
Case $n=5$:
Let $b_1\le b_2\le b_3\le b_4\le b_5$ be the rearrangement of $a_1, a_2, a_3, a_4, a_5$. We have $$a_1^2a_2^2 + a_2^2a_3^2 + a_3^2a_4^2 + a_4^2a_5^2 + a_5^2a_1^2 \le b_1^2b_2^2 + b_1^2b_3^2 + b_2^2b_4^2 + b_3^2b_5^2 + b_4^2b_5^2.$$ See: farfuridi@AoPS, and the first solution of Putnam 1996 B3.
It suffices to prove that \begin{align} & (b_1^3 + b_2^3 + b_3^3 + b_4^3 + b_5^3)^2 \\ \ge \ & (b_1^2 + b_2^2 + b_3^2 + b_4^2 + b_5^2) (b_1^2b_2^2 + b_1^2b_3^2 + b_2^2b_4^2 + b_3^2b_5^2 + b_4^2b_5^2). \end{align}
The Buffalo Way (BW) kills it.
Hint: Let $b_2 = b_1 + s, b_3 = b_2 + t, b_4 = b_3 + r, b_5 = b_4 + u$ for $s, t, r, u\ge 0$.
Case $n=6$:
Let $b_1\le b_2\le b_3\le b_4\le b_5 \le b_6$ be the rearrangement of $a_1, a_2, a_3, a_4, a_5, a_6$. We have $$a_1^2a_2^2 + a_2^2a_3^2 + a_3^2a_4^2 + a_4^2a_5^2 + a_5^2a_6^2 + a_6^2a_1^2 \le b_1^2b_2^2 + b_1^2b_3^2 + b_2^2b_4^2 + b_3^2b_5^2 + b_4^2b_6^2 + b_5^2b_6^2.$$
It suffices to prove that \begin{align} &(b_1^3 + b_2^3 + b_3^3 + b_4^3 + b_5^3 + b_6^3)^2\\ \ge \ & (b_1^2 + b_2^2 + b_3^2 + b_4^2 + b_5^2 + b_6^2) (b_1^2b_2^2 + b_1^2b_3^2 + b_2^2b_4^2 + b_3^2b_5^2 + b_4^2b_6^2 + b_5^2b_6^2). \end{align}
The Buffalo Way (BW) works.