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?

  • 3
    $\begingroup$ 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 $\epsilon_i$ $\endgroup$
    – Deane Yang
    Commented Feb 20, 2013 at 17:15

8 Answers 8


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.

See http://en.wikipedia.org/wiki/Generalized_mean#Generalized_mean_inequality


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.

  • 1
    $\begingroup$ Thanks, it's a very interesting answer even though, as you observed, not exactly what I was looking for. $\endgroup$
    – Vincenzo
    Commented Feb 21, 2013 at 13:15

A result due to Cartwright and Field gives an upper bound for $AM-GM$ of the form you seek:

$$AM-GM \le \frac{1}{2n \min a_i} \sum_{i=1}^{n} (a_i-AM)^2$$ A very naive computation shows that the RHS is $\le \frac{1}{2 \min a_i} (\max a_i - \min a_i)^2$.

This estimate is already better than the suggested naive bound $\max a_i - \min a_i$ when $\max a_i$ and $\min a_i$ are relatively close, specifically: when $\max a_i \le 3 \min a_i$.

It is also worth mentioning that this upper bound was further improved by several authors.


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


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

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

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

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 ratio

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

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

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

$$\frac{s(a_1,\dotsc,a_n)^n}{p(a_1,\dotsc,a_n)}= \frac{s_0^n}{p_0}= F_n(\rho)= F_n\left( \frac{p_0^{n-1}}{\Delta_\max(s_0,p_0)}\right) \geq F_n\left(\frac{p(a_1,\dotsc, a_n)^{n-1}}{\Delta(a_1,\dotsc, a_n)}\right). $$

Here are a few more things things about 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 explicit and strictly increasing polynomial such that $P_n(0)=0$. This implies the sharper inequality

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

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.


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


Upper and lower bounds on on the difference between the arithmetic and geometric means were given in the recent paper available at Bull. Austr. Math. Soc. or arXiv; these bounds are exact in their own terms. Also, see bibliography there.

In particular, the mentioned upper bound on $\frac1n\,\sum_1^n a_i-(a_1\cdots a_n)^{1/n}$ is $$\max\Big[\frac2n\sum_1^n(b_i-\overline b)^2,\frac1n\sum_1^n(b_i-b_\min)^2\Big], $$ where $b_i:=\sqrt{a_i}$, $\overline b:=\frac1n\,\sum_1^n b_i$, and $b_\min:=\min_{1\le i\le n} b_i$.

  • $\begingroup$ I have specified the exact upper bound on the difference AM $-$ GM. $\endgroup$ Commented Sep 13, 2015 at 3:04

The left and right sides are both continuous functions.


$|\exp(\dfrac{1}{n}\ln(a_1)+...+\dfrac{1}{n}\ln(a_n))-AM|\leq \max(a_i) /2\times |\dfrac{1}{n}\sum\limits_{i=1}^n (\ln(a_i))^2-(\dfrac{1}{n}\sum\limits_{i=1}^n \ln(a_i))^2|$

Because $\max(a_i)/2\times x^2-\exp(x)$ and $\max(a_i) /2\times x^2+\exp(x) $ are convexes on $[\min(\ln(a_i)), \max(\ln(a_i))] $


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