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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 another list of $n$ 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 (or really just the arithmetic-geometric mean inequality) $A_1 \geq \sqrt[n]{A_n}$ and we will bound $A_1 - \sqrt[n]{A_n}$ from above in terms of $a_1 - a_n$ by bounding $A_1$ from above and $\sqrt[n]{A_n}$ from below using just $a_1$ and $a_n$. To bound $A_1$ from above,
$$ A_1 = \frac{a_1 + \cdots + a_n}{n} \leq \frac{(n-1)a_1 + a_n}{n} = a_1 - \frac{a_1 - a_n}{n} $$ and to bound $A_n$ from below we write $A_n = a_1\cdots a_n \geq a_n^n$, so $$ \sqrt[n]{A_n} \geq a_n. $$ Therefore $$ 0 \leq A_1 - \sqrt[n]{A_n} \leq \left(a_1 - \frac{a_1 - a_n}{n}\right) - a_n = \left(1 - \frac{1}{n}\right)(a_1 - a_n). $$ Start from 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 $a_i^{(0)} = a_i$. Let's look at the sequence of first coordinates $a_1^{(k)}$. An arithmetic mean of positive numbers is bounded above by the largest number, so $a_1 = a_1^{(0)} \geq a_1^{(1)} \geq a_1^{(2)} \geq \cdots > 0$. Therefore 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 of last coordinates $a_n^{(0)},a_n^{(1)},a_n^{(2)},\dots$ converges to the limit of the sequence of first coordinate $a_1^{(0)}, a_1^{(1)}, a_1^{(2)},\dots$. Since $a_1^{(k)} \geq a_i^{(k)} \geq a_n^{(k)}$, each intermediate sequence $a_i^{(0)},a_i^{(1)},a_i^{(2)},\dots$ converges to the same limit. ]

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 another list of $n$ 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 below to my answer. Order the numbers $a_1,\dots,a_n$ so that $a_1 \geq \cdots \geq a_n > 0$. By Maclaurin's inequality (or really just the arithmetic-geometric mean inequality) $A_1 \geq \sqrt[n]{A_n}$ and we will bound $A_1 - \sqrt[n]{A_n}$ from above in terms of $a_1 - a_n$ by bounding $A_1$ from above and $\sqrt[n]{A_n}$ from below using just $a_1$ and $a_n$. To bound $A_1$ from above,
$$ A_1 = \frac{a_1 + \cdots + a_n}{n} \leq \frac{(n-1)a_1 + a_n}{n} = a_1 - \frac{a_1 - a_n}{n} $$ and to bound $A_n$ from below we write $A_n = a_1\cdots a_n \geq a_n^n$, so $$ \sqrt[n]{A_n} \geq a_n. $$ Therefore $$ 0 \leq A_1 - \sqrt[n]{A_n} \leq \left(a_1 - \frac{a_1 - a_n}{n}\right) - a_n = \left(1 - \frac{1}{n}\right)(a_1 - a_n). $$ Start from an $n$-tuple $(a_1,a_2,\dots,a_n)$ which is ordered so that $a_1 \geq \cdots \geq a_n > 0$ and construct the $n$-tuple $(A_1,\sqrt{A_2},\dots,\sqrt[n]{A_n})$ and keep repeating this, which produces a sequence of $n$-tuples $(a_1^{(k)},a_2^{(k)},\dots,a_n^{(k)})$ for $k = 0,1,2,\dots$, where $a_i^{(0)} = a_i$. Let's look at the sequence of first coordinates $a_1^{(k)}$. An arithmetic mean of positive numbers is bounded above by the largest number, so $a_1 = a_1^{(0)} \geq a_1^{(1)} \geq a_1^{(2)} \geq \cdots > 0$. Therefore the sequence $a_1^{(k)}$ converges as $k \rightarrow \infty$. (The limit is positive because the sequence of last coordinates $a_n^{(k)}$ is non-decreasing and $a_1^{(k)} \geq a_n^{(k)} \geq a_n^{(0)} = a_n$ for all $k$.) 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 of last coordinates $a_n^{(0)},a_n^{(1)},a_n^{(2)},\dots$ converges to the limit of the sequence of first coordinates $a_1^{(0)}, a_1^{(1)}, a_1^{(2)},\dots$. Since $a_1^{(k)} \geq a_i^{(k)} \geq a_n^{(k)}$, each intermediate sequence $a_i^{(0)},a_i^{(1)},a_i^{(2)},\dots$ converges to the same limit. ]

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 another list of $n$ 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 (or really just the arithmetic-geometric mean inequality) $A_1 \geq \sqrt[n]{A_n}$ and we will 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. ]

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 another list of $n$ 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 (or really just the arithmetic-geometric mean inequality) $A_1 \geq \sqrt[n]{A_n}$ and we will bound $A_1 - \sqrt[n]{A_n}$ from above in terms of $a_1 - a_n$ by bounding $A_1$ from above and $\sqrt[n]{A_n}$ from below using just $a_1$ and $a_n$. To bound $A_1$ from above, 
$$ A_1 = \frac{a_1 + \cdots + a_n}{n} \leq \frac{(n-1)a_1 + a_n}{n} = a_1 - \frac{a_1 - a_n}{n} $$ and to bound $A_n$ from below we write $A_n = a_1\cdots a_n \geq a_n^n$, so $$ \sqrt[n]{A_n} \geq a_n. $$ Therefore $$ 0 \leq A_1 - \sqrt[n]{A_n} \leq \left(a_1 - \frac{a_1 - a_n}{n}\right) - a_n = \left(1 - \frac{1}{n}\right)(a_1 - a_n). $$ Start from 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 $a_i^{(0)} = a_i$. Let's look at the sequence of first coordinates $a_1^{(k)}$. An arithmetic mean of positive numbers is bounded above by the largest number, so $a_1 = a_1^{(0)} \geq a_1^{(1)} \geq a_1^{(2)} \geq \cdots > 0$. Therefore 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 of last coordinates $a_n^{(0)},a_n^{(1)},a_n^{(2)},\dots$ converges to the limit of the sequence of first coordinate $a_1^{(0)}, a_1^{(1)}, a_1^{(2)},\dots$. Since $a_1^{(k)} \geq a_i^{(k)} \geq a_n^{(k)}$, each intermediate sequence $a_i^{(0)},a_i^{(1)},a_i^{(2)},\dots$ converges to the same limit. ]

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. ]

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 another list of $n$ 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 (or really just the arithmetic-geometric mean inequality) $A_1 \geq \sqrt[n]{A_n}$ and we will 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. ]