Elementary Symmetric Means as Quasi-Arithmetic Means The elementary symmetric polynomial of degree $k$ in $n$ variables is
$$e_{n,k}(\mathbf{x})=\sum_{1 \leq i_1 \leq i_2 \leq \dots \leq i_k \leq n} x_{i_1} x_{i_2} \cdots x_{i_k} $$
and for positive data $\mathbf{x}$, the corresponding elementary symmetric mean is
$$s_{n,k}(\mathbf{x}):=\sqrt[k]{\frac{e_{n,k}(\mathbf{x})}{\binom{n}{k}}} .$$
Quasi-arithmetic means are defined in terms of an invertible function $f$, and the arithmetic mean $A$
$$M_f(\mathbf{x}) := (f^{-1} \circ A \circ f)(\mathbf{x}) = f^{-1} \left( \frac{f(x_1)+f(x_2)+\cdots+f(x_n)}{n} \right)   .$$
For any $n$, the elementary symmetric mean in $n$ variables of degree $n$ is the geometric mean, which is a quasi-arithmetic mean:
$$s_{n,n}(x_1,\dots,x_n)=\sqrt[n]{x_1x_2\cdots x_n} = \exp\left(\frac{\log(x_1)+\log(x_2)+\cdots+\log(x_n)}{n}\right). $$
My question is: are the other elementary symmetric means (with $k<n$) also quasi-arithmetic? If so, can the conjugating functions $f_{n,k}$ be described explicitly?
My attempt: Obviously $s_{n,1}=A$ is quasi-arithmetic, so the first nontrivial case is $s_{3,2}$, with the functional equation
$$f_{3,2} \circ s_{3,2} = A \circ f_{3,2}, $$ or
$$f_{3,2} \left( \sqrt{ \frac{x_1 x_2+x_1 x_3+x_2 x_3}{3}} \right) = \frac{f_{3,2}(x_1)+f_{3,2}(x_2)+f_{3,2}(x_3)}{3}. $$
I tried pre-composing $f$ with a logarithm and differentiation, but it didn't seem to get me anywhere.
I would greatly appreciate any help on this. Thank you.
 A: If you require that $f$ is, say, $C^1$, then differentiating $M_f$ with respect to any $x_i$ gives
$$\frac{\partial}{\partial x_i} M_f = \frac{1}{f' \left( \frac{\sum f(x_i)}{n} \right)} \frac{f'(x_i)}{n}$$
which means that for any $i \neq j$ we have
$$\frac{ \frac{\partial}{\partial x_i} M_f }{ \frac{\partial}{\partial x_j} M_f } = \frac{f'(x_i)}{f'(x_j)}.$$
So we can check whether $s_{3, 2}$ has this property. Actually it will be slightly more convenient for the purposes of calculating derivatives to check this property after conjugating $s_{3, 2}$ by $f(x) = x^2$ to remove the outer square root, giving a modified mean
$$t_{3, 2}(x_1, x_2, x_3) = \frac{\sqrt{x_1 x_2} + \sqrt{x_2 x_3} + \sqrt{x_3 x_1}}{3}.$$
(Conjugating preserves the property of being quasi-arithmetic so this is fine.) We get
$$\frac{\partial}{\partial x_1} t_{3, 2} = \frac{ \sqrt{x_2} + \sqrt{x_3} }{6 \sqrt{x_1}}$$
and similarly for $x_2, x_3$, which gives
$$\frac{ \frac{\partial}{\partial x_1} t_{3, 2} }{ \frac{\partial}{\partial x_2} t_{3, 2} } = \frac{ (\sqrt{x_2} + \sqrt{x_3}) \sqrt{x_2} }{ (\sqrt{x_1} + \sqrt{x_3}) \sqrt{x_1} }.$$
In particular, this quotient depends nontrivially on $x_3$, so it's not of the form $\frac{f'(x_1)}{f'(x_2)}$ for any function $f$. A similar but more annoying calculation can be done for the other elementary symmetric means. Intuitively this is saying that the elementary symmetric means "mix" the $x_i$ too much to be quasi-arithmetic.
