You want Maclaurin's inequality.  Given $n$ positive numbers $a_1, a_2,\dots,a_n$, 
write 
$$
(x+a_1)(x+a_2)\cdots(x+a_n) = x^n + S_1x^{n-1} + \cdots + S_{n-1}x + S_n, 
$$
so $S_i$ is the $i$-th elementary symmetric function of $a_1,\dots,a_n$. 
For $i = 1,\dots,n$, set $A_i = S_i/\binom{n}{i}$.  When 
$n = 2$, $A_1 = (a_1+a_2)/2$ and $A_2 = a_1a_2$.  Maclaurin's inequality is that
$$
A_1 \geq \sqrt{A_2} \geq \sqrt[3]{A_3} \geq \cdots \geq \sqrt[n]{A_n},
$$
where the inequality signs are all strict unless $a_1,\dots,a_n$ are all equal. 
The inequality of the outer terms, $A_1 \geq \sqrt[n]{A_n}$, is the arithmetic-geometric mean inequality for $n$ positive numbers.

From a list of $n$ positive numbers $a_1,\dots,a_n$ we have produced a list of $n$ new positive numbers $A_1,\sqrt{A_2},\dots,\sqrt[n]{A_n}$. The construction can be repeated. 

Theorem: All the terms in the list tend to the same limit.

Off the top of my head I can't recall a reference where this is proved. 
It was studied by Meissel in 1875 for $n = 3$.

For example, if we start with the three numbers 1, 2, 3 then 
after 4 iterations the three numbers we get all look like 1.9099262335 to 10 digits after the decimal point. 

[Edit: Here is a proof of the common limit, based on Will Sawin's first comment to my answer below. Order the numbers $a_1,\dots,a_n$ so that $a_1 \geq \cdots \geq a_n > 0$.  By Maclaurin's inequality, 
$A_1 \geq \sqrt[n]{A_n}$ and we want to bound 
$A_1 - \sqrt[n]{A_n}$ from above in terms of $a_1 - a_n$.  Well, 
$$
A_1 = \frac{a_1 + \cdots + a_n}{n} \leq \frac{(n-1)a_1 + a_n}{n} = a_1 - \frac{a_1 - a_n}{n} 
$$
and since $A_n = a_1\cdots a_n \geq a_n^n$ we have 
$$
\sqrt[n]{A_n} \geq a_n.
$$
Therefore 
$$
0 \leq A_1 - \sqrt[n]{A_n} \leq \left(a_1 - \frac{a_1 - a_n}{n}\right) - a_n = \left(1 - \frac{1}{n}\right)(a_1 - a_n).
$$
Now start with a sequence $(a_1,a_2,\dots,a_n)$ which is ordered so that $a_1 \geq \cdots \geq a_n > 0$ and construct the sequence $(A_1,\sqrt{A_2},\dots,\sqrt[n]{A_n})$ and keep repeating this, 
which produces a sequence $(a_1^{(k)},a_2^{(k)},\dots,a_n^{(k)})$ for $k = 0,1,2,\dots$, where $a_i^{(0)} = a_i$.  Since an arithmetic mean of positive numbers is bounded above by the largest number, $a_1 = a_1^{(0)} \geq a_1^{(1)} \geq a_1^{(2)} \geq \cdots \geq 0$, so 
the sequence $a_1^{(k)}$ converges as $k \rightarrow \infty$. The above calculation shows 
$$
0 \leq a_1^{(k)} - a_n^{(k)} \leq \left(1 - \frac{1}{n}\right)(a_1^{(k-1)} - a_n^{(k-1)}), 
$$
so $0 \leq a_1^{(k)} - a_n^{(k)} \leq (1 - 1/n)^k(a_1 - a_n)$. Letting $k \rightarrow \infty$ we see the sequence $a_n^{(0)},a_n^{(1)},a_n^{(2)},\dots$ converges to the limit of the sequence $a_1^{(0)}, a_1^{(1)}, a_1^{(2)},\dots$. Therefore each intermediate sequence $a_i^{(0)},a_i^{(1)},a_i^{(2)},\dots$ converges to the same limit. 
]