MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Almost-converses to the AM-GM inequality

Let us consider the Arithmetic Mean -- Geometric Mean inequality for nonnegative real numbers:

$$GM := (a_1 a_2 \ldots a_n)^{1/n} \le \frac{1}{n} \left( a_1 + a_2 + \ldots + a_n \right) =: AM.$$

It is known that the converse inequality ($\ge$) holds if and only if all the $a_i$'s are the same.

Therefore, we can expect that if the $a_i$'s are almost the same, then a converse inequality almost holds. For example, we may look for an inequality of the form $AM \le GM + f(\Delta,n)$ where $\Delta$ is the ratio between $\max_i a_i$ and $\min_i a_i$, but this is just one possibility.

Are there any natural ways to formalize the above intuition?

-
 I'm sure this has been done before, but you can try to work something out yourself by just setting each $a_i = a + \epsilon_i$ (or $a_i = (1 + \epsilon_i)a$ and expanding the left side up to first order in the $\epsilon_i$ and an error term that is quadratic in $\epsilon_i$ – Deane Yang Feb 20 at 17:15

Power mean inequality can give many bounds for the difference between AM and GM. Most simple is $$AM - GM \leq \max_i a_i - \min_i a_i.$$ Another bound is $$AM - GM \leq AM - HM = \frac{a_1+\dots+a_n}{n} - \frac{n}{1/a_1 + \dots + 1/a_n}$$ etc.

-

It's not precisely what you asked about, but this paper by Gluskin and Milman shows that, for "most" sequences $a_1, \dotsc, a_n$, the AM-GM inequality can be reversed up to a multiplicative constant. The paper contains a number of observations which come closer to directly addressing your question.

-
 Thanks, it's a very interesting answer even though, as you observed, not exactly what I was looking for. – Vincenzo Feb 21 at 13:15

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.

-

The left and right sides are both continuous functions.

-

Proposition 1 in this paper might be what you are looking for.

-