Calculate the Mean mean of two numbers disclaimer: the Mean mean is made up by me
Define the arithmetic mean of \$n\$ numbers as
$$M_1(x_1,...,x_n)=\frac{x_1+x_2+...+x_n}{n}$$
Define the geometric mean of \$n\$ numbers as
$$M_0(x_1,...,x_n)=\root{n}\of{x_1x_2...x_n}$$
Define the harmonic mean of \$n\$ numbers as
$$M_{-1}(x_1,...,x_n)=\frac{n}{\frac{1}{x_2} + \frac{1}{x_2} + ... + \frac{1}{x_n}}$$
Define the quadratic mean of \$n\$ numbers as
$$M_2(x_1,...,x_n)=\root\of{\frac{x_1^2+x_2^2+...+x_n^2}{n}}$$
The Mean mean (\$M_M\$) is defined as follows:
Define four sequences (\$a_k, b_k, c_k, d_k\$) as $$a_0=M_1(x_1,...,x_n),\\b_0=M_0(x_1,...,x_n),\\c_0=M_{-1}(x_1,...,x_n),\\d_0=M_2(x_1,...,x_n),\\
a_{k+1}=M_1(a_k,b_k,c_k,d_k),\\b_{k+1}=M_0(a_k,b_k,c_k,d_k),\\c_{k+1}=M_{-1}(a_k,b_k,c_k,d_k),\\d_{k+1}=M_2(a_k,b_k,c_k,d_k)$$
All four sequences converge to the same number, \$M_M(x_1,x_2,...,x_n)\$.
Example
The Mean mean of 1 and 2 is calculated as follows: start with $$a_0 = (1+2)/2 = 1.5, b_0 = \root\of{1 * 2} = \root\of2 \approx 1.4142,\\ c_0 = \frac2{\frac1{1}+\frac1{2}} = \frac4{3} \approx 1.3333, d_0 = \root\of{\frac{1^2+2^2}2} = \root\of{\frac5{2}} \approx 1.5811.$$
Then
$$a_1 = \frac{1.5+1.4142+1.3333+1.5811}4 \approx 1.4571,\\ b_1 = \root^4\of{1.5*1.4142*1.3333*1.5811} \approx 1.4542,\\ c_1 = \frac4{\frac1{1.5}+\frac1{1.4142}+\frac1{1.3333}+\frac1{1.5811}} \approx 1.4512,\\ d_1 = \root\of{\frac{1.5^2+1.4142^2+1.3333^2+1.5811^2}4} \approx 1.4601.$$
The further calculation of the sequences should be clear. It can be seen that they converge to the same number, approximately \$1.45568889\$.
Challenge
Given two positive real numbers, \$a\$ and \$b\$ (\$a&lt;b\$), calculate their Mean mean \$M_M(a,b)\$.
Test cases
1 1 =&gt; 1
1 2 =&gt; 1.45568889
100 200 =&gt; 145.568889
2.71 3.14 =&gt; 2.92103713
0.57 1.78 =&gt; 1.0848205
1.61 2.41 =&gt; 1.98965438
0.01 100 =&gt; 6.7483058

Notes

Your program is valid if the difference between its output and the correct output is not greater than 1/100000 of the absolute value of the difference between input numbers.
The output should be a single number.

This is code-golf, so the shortest code wins!
Wolfram Language (Mathematica), 52 bytes

#//.x_:&gt;N@{M@x,E^M@Log@x,1/M[1/x],M[x^2]^.5}&amp;
M=Mean


Try it online!

In my first approach I used these builtins
Mean GeometricMean  HarmonicMean and RootMeanSquare  

Here are some substitutions for saving bytes  

HarmonicMean -->  1/Mean[1/x]  by @Robin Ryder (3 bytes saved)
GeometricMean --> E^Mean@Log@x by @A. Rex (2 bytes saved)
RootMeanSquare --> Mean[x^2]^.5 by @A. Rex (4 bytes saved)   

finally we can assign Mean to M (as proposed by @ovs) and save 5 more bytes