A generalization of arithmetic and geometric means using elementary symmetric polynomials Let $a_1, a_2, \ldots, a_n$ be positive real numbers.  A while ago I noticed that if you form the polynomial
$$
P(x) = (x - a_1)(x-a_2) \cdots (x-a_n)
$$
then:


*

*The arithmetic mean of $a_1, \ldots, a_n$ is the positive number $m$ such that $(x - m)^n$ and $P(x)$ have the same coefficient of $x^{n-1}$.

*The geometric mean of $a_1, \ldots, a_n$ is the positive number $m$ such that $(x - m)^n$ and $P(x)$ have the same coefficient of $x^{0}.$


It looks like this can be extended: for any $0 \le i \le n-1$, let the $i$th mean be the number $m_i$ such that $(x - m_i)^n$ and $P(x)$ have the same coefficient of $x^i$.  (Alternatively, one can define $m_i$ in terms of elementary symmetric polynomials.)
For instance, with three variables $x, y, z$ we get
\begin{align*}
m_0 &= \sqrt[3]{xyz} \\
m_1 &= \sqrt{\frac{xy + yz + zx}{3}} \\
m_2 &= \frac{x + y + z}{3}
\end{align*}
I am not sure what qualifies something as a "mean", but $m_1$ is symmetric and lies strictly between the min and the max, and it probably has other properties as well.  Here's a more concrete question:

Question: Must it be true that $m_0 \le m_1 \le m_2 \le \cdots \le m_{n-1}$?


Denote, $(\overline{a}) = (a_1,\cdots,a_n)$ and $\displaystyle P(x) = \prod\limits_{k=1}^{n}(x - a_k) = x^n +\sum\limits_{k=1}^{n} (-1)^k\binom{n}{k}u_k(\overline{a})x^{n-k}$
where, $\displaystyle u_k(\overline{a}) = \dfrac{\sum\limits_{1\le j_1<\cdots< j_k \le n}a_{j_1}\cdots a_{j_k}}{\binom{n}{k}} = m_{n-k}^{k}$ (in your notation).
Note that $P(x)$ has $n$ real roots in the interval $\left[\min\limits_{i=1}^n\{a_i\},\max\limits_{i=1}^n\{a_i\}\right]$, 
Thus $P^{(n-2)}(x) = \dfrac{n!}{2}(x^2 - 2u_1x + u_2)$ has two real roots in that interval, i.e., $u_1^2 \ge u_2$.
Similarly apply the same idea for the $(n-2)^{th}$ derivative of the polynomial with roots $\dfrac{1}{a_k}$, ($k = 1(1)n$) 
We get, $\displaystyle \frac{1}{\binom{n}{2}}\sum\limits_{i<j}a_ia_j \le \frac{1}{\binom{n}{1}^2}\left(\sum\limits_{i=1}^{n} \frac{1}{a_i}\right)^2 \implies u_{n-1}^2 \ge u_nu_{n-2}$.
We show, $u_{k-1}(\overline{a})u_{k+1}(\overline{a}) \le u_k^2(\overline{a})$ for $k = 2,3,\cdots,n-1$,
We can prove this result by induction on $n$, suppose the inequality holds for any $n-1$ positive real numbers.
We have $\displaystyle P'(x) = n\left(x^{n-1} + \sum\limits_{k=1}^{n-1}(-1)^k\binom{n-1}{k}u_k(\overline{a})x^{n-k-1}\right)$
If the roots of $P'(x)$ are $b_k$, for $k=1,2,\cdots,n-1$, (which, are positive reals by M.V.T.).
and define $v_k = \dfrac{\sum\limits_{1\le j_1<\cdots< j_k \le n-1}b_{j_1}\cdots b_{j_k}}{\binom{n-1}{k}}$
Then $\displaystyle P'(x) = n\prod\limits_{k=1}^{n-1}(x - b_k) = n\left(x^{n-1} + \sum\limits_{k=1}^{n-1}(-1)^k\binom{n-1}{k}v_kx^{n-k-1}\right)$
Thus, $u_k = v_k$ for $k=1,2,\cdots,n-1$ and by the induction hypothesis on the numbers $(b_1,\cdots,b_{n-1})$ we get $u_k^2 \ge u_{k+1}u_{k-1}$ for $k = 2,\cdots,n-2$ and together with $u_{n-1}^2 \ge u_nu_{n-2}$ completes the induction.
If we take the $k$th inequality to the $k$th power and then multiply all these inequalities for $k=1,\cdots,r$, we get,
$u_1^2 u_2^4 u_3^6 \cdots u_{r-1}^{2r-2}u_{r}^{r-1}u_{r+1}^r \le u_1^2u_2^4\cdots u_r^{2r} \implies u_{r+1}^r \le u_{r}^{r+1}$ for $r=1, \cdots,n-1$
Thus, $u_r^{1/r} = m_{n-r}$ forms a non increasing sequence.
