Let $e_k(x_1,\cdots, x_n) := \sum_{1\leq j_1 < \cdots < j_k \leq n}x_{j_1}\cdots x_{j_k}$ be elementary symmetric polynomials (see e.g. https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial). For any fixed $0<x_1\leq x_2 \leq \cdots \leq x_n \in \mathbb{R}$ let's define the following: \begin{equation} s_k := \frac{1}{k}\left(1 + \frac{e_{n-k+1}(x_1,\cdots,x_n)}{e_{n-k}(x_1,\cdots,x_n)}\right) \end{equation} for $k = 1,\cdots,n$. I would like to know whether the following is true:
There exists no $k = 2,\cdots,n-1$ such that $s_k > s_{k-1}$ and $s_k > s_{k+1}$. (Prove or Disprove)
In other words, the sequence $s_k$ has no 'local maximum', for any given fixed $0<x_1\leq x_2 \leq \cdots \leq x_n$. I could verify in the simplest case where $x_1 = x_2 = \cdots = x_n =: x$ that this statement holds, we have $e_k(x,\cdots,x) = (n!/k!(n-k)!)x^k$, and so \begin{equation} s_k = \frac{1}{k}\left(1 + \frac{(n-k)!k!}{(n-k+1)!(k-1)!}x\right) = \frac{1}{k}\left(1 + \frac{k}{n-k+1}x\right) = \frac{1}{k} + \frac{x}{n-k+1}. \end{equation} In this case, for any $k = 2,\cdots,n-1$, we will either have $s_{k-1} \leq s_k \leq s_{k+1}$, or $s_{k-1} \geq s_k \geq s_{k+1}$, or $s_k \leq s_{k-1}, s_{k+1}$, but not $s_k > s_{k-1}, s_{k+1}$.
I'm not sure how to prove (or disprove) the statement in general, but I numerically compute $s_k$ for randomized $0 < x_1 \leq x_2 \leq \cdots \leq x_n$ many times and the statement appears to be true so far. Below is the plot of an example computed $s_k$ with $n = 20$ for a sample set of $0<x_1\leq x_2\leq \cdots\leq x_n$:
as you can see there is no local maximum. If someone could help or suggest ways to either prove or disprove the statement I would really be appreciated!