Computing cube roots using three number means I've asked a question some time ago about Computing square roots with arithmetic-harmonic mean but it turned out that this method is exactly the same as Newton's method (or Babylonian method) for square roots.
Now I figured out how to compute cube roots in the same way - and in the case of cube roots this method is apparently not the same as Newtons' method.
Let's introduce the following algorithm for some $x,y,z$:
$$a_0=\frac{x+y+z}{3},~~~~~~~b_0=\frac{xy+yz+zx}{x+y+z},~~~~~~~c_0=\frac{3xyz}{xy+yz+zx}$$
$$a_{n+1}=\frac{a_n+b_n+c_n}{3},~~~~~~~b_{n+1}=\frac{a_nb_n+b_nc_n+c_na_n}{a_n+b_n+c_n},~~~~~~~c_{n+1}=\frac{3a_nb_nc_n}{a_nb_n+b_nc_n+c_na_n}$$
We can obviously see that:
$$a_{n+1}b_{n+1}c_{n+1}=a_nb_nc_n=xyz$$
Thus, if the algorithm converges, we have:
$$\lim_{n \to \infty}a_n=\lim_{n \to \infty}b_n=\lim_{n \to \infty}c_n=\sqrt[3]{xyz}$$
For practical purposes it is much better to take $b_n$ as an estimate of the root, as can be shown by the following:
$$x=2,~~~~y=z=1$$
$$\begin{array}( n & a_n & b_n & c_n & \sqrt[3]{xyz} \\ 0 & 1.3333333333 & \color{blue}{1.25} & 1.2 & 1.2599210499 \\ 1 & 1.2611111111 & \color{blue}{1.2599}118943 & 1.2587412587 & 1.2599210499 \\ 2 & 1.2599214214 & \color{blue}{1.2599210499} & 1.2599206784 & 1.2599210499  \end{array}$$

This algorithm can be generalized for $n$th roots as well, but it will involve a lot more calculations at each step, since we will have to compute $n$ means from sums of products of $k$-tuples with $k=(1,2,\dots,n)$.

How do we prove this algorithm converges for any $x,y,z$?
Is it really different from Newton's method? If it is, could it be more efficient in some cases?

 A: Let $x,y,z> 0$. It is easy to see directly from the recurrence that $a_{n}\ge b_{n}\ge c_{n}>0$,$~n\ge 0$. Therefore $a_{n+1}\le a_n$ and $c_{n}\le c_{n+1}$.
As a result one gets the following chain of inequalities
$$
c_0\le c_1\le...\le c_n\le  b_n\le a_n\le...\le a_1\le a_0.
$$
The sequence $c_n$ is monotone increasing and bounded from above, so it has a limit $\lim_{n\to\infty} c_n=c$. Similarly the sequence $a_n$ is monotone decreasing and bounded from below, so it has a limit $\lim_{n\to\infty} a_n=a$.
Since $a_nb_nc_n=xyz$, and $a_n$ and $c_n$ has a limit, so does $$\lim_{n\to\infty} b_n=\lim_{n\to\infty}\frac{xyz}{a_nc_n}=\frac{xyz}{ac}=b,\quad c\le b\le a.\tag{*}$$
Now one can take the limit of the equation $b_{n+1}=\frac{a_nb_n+b_nc_n+c_na_n}{a_n+b_n+c_n}$:
$$
b=\frac{ab+bc+ca}{a+b+c}\implies b^2=ac.\tag{**}
$$
Combining $(*)$ and $(**)$ one obtains
$$
b=\sqrt[3]{xyz}.
$$
The limit of $a_{n+1}=\frac{a_n+b_n+c_n}{3}$
$$
2a=b+c,
$$
together with inequality $c\le b\le a$ implies that $a=b=c$.
