<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] Let $\mathbf{x}$ denote a vector of positive real numbers (you can take it for your purposes to be 3 such numbers). Let $\mathbf{w}$ denote a vector of positive real numbers of the same length (i.e. 3-dimensional in this case), with $\sum_i w_i=1$; the simplest option is all $w_i$ being equal. Define $M_p:=\left(\sum_i w_i x_i^p\right)^{1/p}$ for $p\neq 0$. We can define $M_0$ by continuity; it's $\prod_i x_i^{w_i}$. Then $M_p$ is a strictly increasing function of $p$. If all $w_i$ are equal, $M_{-1}$ is the HM, $M_0$ is the GM, $M_1$ is the AM etc. Thus $M_p>M_q$ for $p>q$ is called the power means inequality. If you want a greater variety of inequalities, you can use the fact that $M_p\left(\mathbf{x}\right)>M_p\left(\mathbf{y}\right)$ provided each $x_i\geq y_i$ and at least one $x_j>y_j$. In particular, you can replace a vector entry with a power-mean (including geometric) of greater entries to created a complicated "nested" expression. Variations in $\mathbf{w}$ can give you some inequalities too, but only if you assume (for example) that $x_1>x_2$.
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