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. When $n = 2$ this is the arithmetic-geometric mean inequality.
From a list of $n$ positive numbers $a_1,\dots,a_n$ we have produced a list of $n$ new positive numbers $A_1,\sqrt{A_2},\dots,\sqrt[n]{A_n}$. Repeat the construction. When $n = 2$ this is the arithmetic-geometric mean recursion.
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.