# 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 $n\le 16$?

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 $n \le 16$? It is understood that $a_{n+1} = a_1$.

The restriction to $n \le 16$ comes from a known counterexample for $n = 17$ as follows: Arrange the $a_i$ in blocks such that 9 consecutive $a_i =1.1$ and the following 8 consecutive $a_i =1$. Then

$$f(\{a_i\}) = (9\cdot 1.1^{1.5} + 8)^2 - (9\cdot 1.1 + 8)(8\cdot 1.1^2 + 7 + 2 \cdot 1.1) = -0.0097 <0$$

Let me further sketch my thoughts why it may be believed that the inequality indeed holds for $n \le 16$.

First, it can be shown that all block arrangements for $n \le 16$ satisfy the inequality, which can be seen as follows. Let a block of $n\cdot p$ many $a_i =A$ and the remaining $n\cdot (1-p)$ many $a_i =B$, where $0<p<1$. Let $B = x^2 A$. Then we have

$$\frac{1}{A^3}f(\{a_i\}) = (n p + n (1-p) \cdot x^3 )^2 - (np + n (1-p) x^2)(n p -1 + (n (1-p)-1) x^4 + 2 x^2)$$ which can be rearranged to

$$\frac{1}{n A^3} f(\{a_i\}) = - n p(1 - p) \cdot x^2 (x - 1)^2 +(p + x^2(1 - p))\cdot(x^2 - 1)^2$$

Since the leading term with $n$ is always negative and the remaining term is always positive, it is clear that for large $n$, $f <0$, whereas for small $n$, $f >0$. Regarding $n=n_0$ for a moment as a real variable, we can determine the $n_0$ where $f=0$ as

$$n_0 = n_0(p,x) = \frac{(p + x^2(1 - p))\cdot(x^2 - 1)^2}{p(1 - p) \cdot x^2 (x - 1)^2 }$$ Detailed analysis (solving for the minimum) of this function shows that the smallest value will be attained for $p=0.5$ and $x\to 1$. $x=1$ is not allowed since then all $\{a_i\}$ would equal. Now we have $$\lim_{x\to 1} n_0(p=0.5,x) = 16$$ Since $x \ne 1$, we always have $n_0(p,x) > 16$, which shows that all block arrangements with $n \le 16$ satisfy the inequality.

Now if it can be shown that indeed block arrangements constitute the "worst case" of the given inequality, the proof would be complete. I didn't manage to do that, though, or to devise other proof or disproof methods.

Note: the problem originated from this question where it was falsely believed that the inequality would hold for all $n$.

• Just by choosing random numbers I found counterexamples for n=9...16 (unless I made some error). – Martin R Oct 3 '17 at 15:23
• Some counter-examples: n=9: (40, 37, 40, 50, 60, 65, 65, 56, 47), n=10: (36, 40, 44, 48, 75, 100, 100, 94, 70, 54), n=16: (36, 40, 44, 48, 75, 100, 100, 94, 70, 54) – Martin R Oct 3 '17 at 15:58
• @MartinR Thank you for your counterexamples. In the last one, you write $n=16$ buth there are 10 components. Your cases all violate the conjecture, so indeed it doesn't hold. – Andreas Oct 3 '17 at 17:06
• @MartinR How did you come about the counterexamples? Do they give a method how one could fix an $n$ up to where the conjecture holds? $n=3$ and $n=4$ can be proven, how about $n=5,6,7,8$ ? – Andreas Oct 3 '17 at 17:08
• That was a copy/paste error, here is the example for n=16: (78, 76, 47, 35, 25, 33, 39, 45, 61, 67, 77, 78, 70, 90, 92, 78). – I simply ran a computer program that generates random numbers in the range 0...100 :) I got no counter examples for n=5...8 (in reasonable time). – Martin R Oct 3 '17 at 17:18