# Smallest Sum of products, under cyclic permutations

Let $a_i$ be real positive numbers, $i=0, \dots, N-1$, with $N$ even.

Define sums $S_k = \sum_{i=0}^{N-1} a_i a_{i+k}$, for $k=0, \dots, N-1$, where the indices are understood to be cyclic, i.e. $a_{N+j} = a_j$.

Prove or disprove: For all $k$: $S_k \ge S_{N/2}$.

Ideas: one approach to this might be the Rearrangement Inequality (RI). Without loss of generality, we can order the $\{a_i\}$ in such a way that they are ascending. Then amongst the $\{a_{i+k}\}$, only $\{a_{i+0}\}$ is ascending and for all other $k$, the $\{a_{i+k}\}$ are not ascending (as a consequence of the cyclic nature). So by the RI, the largest sum is $S_0$.

In the same vein, if one finds that $S_{N/2}$ is "most disturbing" to the ascending order, then using the RI would show the claim. I didn't manage to do that.

## 1 Answer

The conjecture is false.

Let $N=4$

and $a_0=1,a_1=10,a_2=11,a_3=1000$

Then $S_1=S_3=12120<S_2=20022$

• Nice counterexample! Thanks! – Andreas Sep 27 '17 at 9:30