<s>[INST] Extending the ordered sequence of 'three-number means' beyond AM, GM and HM I want to create an ordered sequence of various 'three-number means' with as many different elements in it as possible.
So far I've got $12$ ($8$ unusual ones are highlighted):
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \color{blue}{  \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}} \geq   $$
$$\geq  \color{blue}{\frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}} } \geq \frac{x+y+z}{3} \geq  $$
$$ \geq \color{blue}{ \frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} } \geq  \color{blue}{\sqrt{\frac{xy+yz+zx}{3}}} \geq $$
$$\geq  \color{blue}{\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{3}} \geq \color{blue}{\sqrt{\frac{x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}}{3}}} \geq $$
$$  \geq \sqrt[3]{xyz} \geq \color{blue}{ \frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}} \geq $$
$$ \geq\color{blue}{ 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}}} \geq   \frac{3xyz}{xy+yz+zx}$$
All the inequalities here are proven (note that $x,y,z>0$).
The rules are as follows - only explicit expressions allowed (no AGM for example) and these expressions should contain only addition, multiplication, division, squares, cubes, square and cube roots.
For any new mean we need to find a place between a pair of existing means, so we can keep our sequence precisely ordered.
I partially placed several other means, for example:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \geq \\  \geq \color{blue}{ \frac{\sqrt[6]{(x^2+y^2)(y^2+z^2)(z^2+x^2)}}{\sqrt{2}}} \geq \frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2}$$
$$\sqrt{\frac{xy+yz+zx}{3}} \geq \color{blue}{\frac{xy+yz+zx}{x+y+z}} \geq \sqrt[3]{xyz}$$
But I can't place them exactly among other means so far.
 [/INST] Suppose $f(x,y,z)$ is a function that is:


*

*symmetric, i.e. $f(x,y,z)=f(y,x,z)=f(x,z,y)$ and so on,

*positive, i.e. $f(x,y,z)\ge0$ for all $x,y,z\ge0$, and

*has power-law scaling with exponent $p$, that is, $f(ax,ay,az)=a^pf(x,y,z)$ for all $a\ge0$.


Suppose $g(x,y,z)$ is a similar function but with exponent $q\ne p$. Then
$$m(x,y,z)=\left(\frac{f(x,y,z)\,g(1,1,1)}{f(1,1,1)\,g(x,y,z)}\right)^{q/p}$$
is a "three-number mean".
(Edit: The OP hasn't provided any criteria for what counts as a "mean", but for one thing we probably want the convex hull property $m(x,y,z)\in[\min(x,y,z),\max(x,y,z)]$, which the above function need not satisfy in general. So it probably shouldn't be called a mean until we figure out additional conditions to impose on $f$ and $g$ to guarantee this.)

If $f(x,y,z)$ is a polynomial, it must be a linear combination of monomial symmetric polynomials of degree $p$, and for positivity it is sufficient (though not necessary) that the linear combination is convex.
Almost all of the means in your list can be constructed this way, with $f$ and $g$ being convex combinations of monomial symmetric polynomials either in $x,y,z$ or in $\sqrt x,\sqrt y,\sqrt z$. The sole exception is the numerator of the second example, $\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}$, which is not a polynomial but is still symmetric, positive, and scales with exponent $p=1$.

In particular, suppose we take $p=2$ and $q=0$. The two monomial symmetric polynomials of degree $2$ are $x^2+y^2+z^2$ and $xy+yz+zx$, so $f(x,y,z)$ is a multiple of $\alpha(x^2+y^2+z^2) + (1-\alpha)(xy+yz+zx)$ for some $\alpha\in[0,1]$. Therefore our mean is
$$m(x,y,z) = \sqrt{\frac{\alpha(x^2+y^2+z^2) + (1-\alpha)(xy+yz+zx)}3}.$$
It includes the following special cases which exist in your sequence:
$$\begin{align}
\alpha &= 1 \implies & m(x,y,z) &= \sqrt{\frac{x^2+y^2+z^2}3}, \\
\alpha &= \tfrac12 \implies & m(x,y,z) &= \frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2\sqrt3}, \\
\alpha &= \tfrac13 \implies & m(x,y,z) &= \frac{x+y+z}3, \\
\alpha &= 0 \implies & m(x,y,z) &= \sqrt{\frac{xy+yz+zx}3}. \\
\end{align}$$
In this way, using different $p$ and $q$, one can construct arbitrarily many continuous families of "three-number means".
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