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Here is an old result of Siegel that is related to your question.

Set

$$s=s(a_1,\dotsc, a_n)=\frac{1}{n} (a_1+\cdots +a_n),$$

$$p= p(a_1,\dotsc, a_n),$$

$$\Delta= \Delta(a_1,\dotsc, a_n)=\prod_{i,j}(a_i-a_j)^2.$$

$$\frac{s^n}{p}\geq 1.$$

Observe that $s$ is homogeneous of degree $1$, $p$ is homogeneous of degree $n$ and $\Delta$ is homogeneous of degree $n(n-1)$ in the variables $a_j$. In particular, the ration

$$R= \frac{p^{n-1}}{\Delta}$$

is homogeneous of degree $0$. Note that $\Delta=0$ when two of the numbers $a_j$ are equal. In particular, large $\Delta$ would mean that the numbers are "far from being equal". Equivalently the larger $\Delta$ is, the more "dispersed" are the numbers $a_j$.

One can ask how dispersed can the numbers $a_j$ be given that $s$ and $p$ are fixed.

In other words we ask to find

$$\max \Delta(a_1,\dotsc, a_n)$$

given that

$$s(a_1,\dotsc, a_n)=s_0,\;\;p(a_1,\dotsc, a_n)=p_0.$$

This constrained maximum exists and can be described explicitly as the discriminant of a certain Laguerre polynomial. I will denote it by $\Delta_\max(s_0,p_0)$.

I will set

$$\rho=\rho(s_0,p_0)= \frac{p_0^{n-1}}{\Delta_\max(s_0,p_0)}.$$

Then there exists an explicit but very complicated strictly decreasing continuous function

$$F_n: (0,\infty)\to (1,\infty)$$

such that

$$\lim_{t\to\infty} F_n(t)=1,$$

$$\frac{s_0^n}{p_0}= F_n(\rho)= F_n\left( \frac{p_0^{n-1}}{\Delta_\max(s_0,p_0)}\right).$$

A few things a bout the function $F_n$. It is described as a composition $Q_n\circ P_n^{-1}$, were

$$Q_n: (0,\infty)\to (1,\infty)$$

is a strictly decreasing very explicit rational function and

$$P_n:(0,\infty)\to (0,\infty)$$

is a very explict and strictly increasing polynomial such that $P_n(0)=0$. This implies the sharper inequality

$$s(a_1, \dotsc, a_n)^n \geq F\left(\frac{p(a_1,\dotsc, a_n)^{n-1}}{\Delta(a_1,\dotsc, a_n)}\right)p(a_1,\dotsc, a_n),$$

with equality iff

$$\Delta(a_1,\dotsc,a_n)=\Delta_\max(s,p).$$

For more details see Sec. 8.6 of the beautiful book Special Functions by G.E. Andrews, R. Askey, R. Roy.

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Here is an old result of Siegel that is related to your question.

Set

$$s=s(a_1,\dotsc, a_n)=\frac{1}{n} (a_1+\cdots +a_n),$$

$$p= p(a_1,\dotsc, a_n),$$

$$\Delta= \Delta(a_1,\dotsc, a_n)=\prod_{i,j}(a_i-a_j)^2.$$

$$\frac{s^n}{p}\geq 1.$$

Observe that $s$ is homogeneous of degree $1$, $p$ is homogeneous of degree $n$ and $\Delta$ is homogeneous of degree $n(n-1)$ in the variables $a_j$. In particular, the ration

$$R= \frac{p^{n-1}}{\Delta}$$

is homogeneous of degree $0$. Note that $\Delta=0$ when two of the numbers $a_j$ are equal. In particular, large $\Delta$ would mean that the numbers are "far from being equal". Equivalently the larger $\Delta$ is, the more "dispersed" are the numbers $a_j$.

One can ask how dispersed can the numbers $a_j$ be given that $s$ and $p$ are fixed.

In other words we ask to find

$$\max \Delta(a_1,\dotsc, a_n)$$

given that

$$s(a_1,\dotsc, a_n)=s_0,\;\;p(a_1,\dotsc, a_n)=p_0.$$

This constrained maximum exists and can be described explicitly as the discriminant of a certain Laguerre polynomial. I will denote it by $\Delta_\max(s_0,p_0)$.

I will set

$$\rho=\rho(s_0,p_0)= \frac{p_0^{n-1}}{\Delta_\max(s_0,p_0)}.$$

Then there exists an explicit but very complicated strictly decreasing continuous function

$$F_n: (0,\infty)\to (1,\infty)$$

such that

$$\lim_{t\to\infty} F_n(t)=1,$$

$$\frac{s_0^n}{p_0}= F_n(\rho)= F_n\left( \frac{p_0^{n-1}}{\Delta_\max(s_0,p_0)}\right).$$

A few things a bout the function $F_n$. It is described as a compostion composition $Q_n\circ P_n^{-1}$, were

$Q_n:$ Q_n: (0,\infty)\to (1,\infty)$1,\infty) $$is a strictly decreasing very explicit rational function and P_m:(0,\infty)\to P_n:(0,\infty)\to (0,\infty) 0,\infty)$$ is a very explict and strictly increasing polynomial such that$P_n(0)=0$. For more details see Sec. 8.6 of the beautiful book Special Functions by G.E. Andrews, R. Askey, R. Roy. 3 added 317 characters in body Here is an old result of Siegel that is related to your question. Set $$s=s(a_1,\dotsc, a_n)=\frac{1}{n} (a_1+\cdots +a_n),$$ $$p= p(a_1,\dotsc, a_n),$$ $$\Delta= \Delta(a_1,\dotsc, a_n)=\prod_{i,j}(a_i-a_j)^2.$$ The AM-GM inequality reads $$\frac{s^n}{p}\geq 1.$$ Observe that$s$is homogeneous of degree$1$,$p$is homogeneous of degree$n$and$\Delta$is homogeneous of degree$n(n-1)$in the variables$a_j$. In particular, the ration $$R= \frac{p^{n-1}}{\Delta}$$ is homogeneous of degree$0$. Note that$\Delta=0$when two of the numbers$a_j$are equal. In particular, large$\Delta $would mean that the numbers are "far from being equal". Equivalently the larger$\Delta$is, the more "dispersed" are the numbers$a_j$. One can ask how dispersed can the numbers$a_j$be given that$s$and$p$are fixed. In other words we ask to find $$\max \Delta(a_1,\dotsc, a_n)$$ given that $$s(a_1,\dotsc, a_n)=s_0,\;\;p(a_1,\dotsc, a_n)=p_0.$$ This constrained maximum exists and is achieved when$a_1,\dotsc, a_n$are can be described explicitly as the zeros discriminant of a certain Laguerre polynomial. I will denote it by$\Delta_\max(s_0,p_0)$. I will set $$\mu=\mu(s_0,p_0)= rho=\rho(s_0,p_0)= \frac{p_0^{n-1}}{\Delta_\max(s_0,p_0)}.$$ Then there exists an explicit but very complicated strictly decreasing continuous function $$F_n: (0,\infty)\to (1,\infty)$$ such that $$\lim_{\mu\to\infty} F_n(\mu)=1\lim_{t\to\infty} F_n(t)=1,$$ $$\frac{s_0^n}{p_0}= F_n(\mu)= F_n(\rho)= F_n\left( \frac{p_0^{n-1}}{\Delta_\max(s_0,p_0)}\right).$$ A few things a bout the function$F_n$. It is described as a compostion$Q_n\circ P_n^{-1}$, were$Q_n: (0,\infty)\to (1,\infty)$is a strictly decreasing very explicit rational function and$P_m:(0,\infty)\to (0,\infty)$is a very explict and strictly increasing polynomial such that$P_n(0)=0\$.

For more details see Sec. 8.6 of the beautiful book Special Functions by G.E. Andrews, R. Askey, R. Roy.

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