# Inequalities involving arithmetic, geometric and harmonic means

Let $A$, $G$ and $H$ denote the arithmetic, geometric and harmonic means of a set of $n$ values. It is well-known that $A$, $G$, and $H$ satisfy $$A \ge G \ge H$$ regardless of the value $n$. Furthermore, for $n=2$ we have $$G^2 = AH$$ By coincidence I found the following result for $n=3$ which I haven't seen before: $$A^2H \ge G^3 \ge AH^2,\qquad n=3$$ I have looked around to find more general inequalities like this for other values of $n$ but I couldn't find any. Are there similar results for the general case, i.e. arbitrary $n$? Any pointer is greatly appreciated.

• – David E Speyer Apr 11 '13 at 14:12
• Thanks David, there's indeed a lot of information there. – Matt L. Apr 11 '13 at 17:02

Edit: I just found Macavity's answer

to the question

math.stackexchange.com/q/803960/204426

uses the development below, preceding me, I am sorry I missed it before. As I found the construction in a different context and have written it up anyway now, I leave this answer here for convenience:

There is a generalization of your inequality to arbitrary $n$. It is derived by combining an elementary generalization of the usual power means, with the well known inequalities of Newton/Maclaurin, which you will find for example in section 2.22 of a classic text of Hardy, Littlewood and Polya, "Inequalities", republished in the Cambridge Mathematical Library series.

Given $n$ $positive$ numbers $a_n$, their geometric mean $M_0$ is defined by

$$M_0 (a_j) \ := \ \sqrt[n]{\prod^n_{j=1} \, a_j} \, ,$$

while their power mean of exponent $r \, \neq \, 0$ is defined as

$$M_r (a_j) \ := \ \left( \frac{1}{n} \sum^n_{j=1} \, a_j^r \right)^{1/r} \, ,$$

so that in your notation $M_1(a_j) \, = \, A,$ $M_{-1}(a_j) \, = \, H$. Lets generalize this by forming the means of exponent $r$ of all distinct products we get when choosing $k$ distinct $a_j$ from our set. In a slight abuse of notation we write $m \, \in \, S_k$ for each such possibility and the corresponding indices, which should not cause confusion. For $r \, = \, 0$ we have:

$$M_{0,k} (a_j) \ := \ \sqrt[{n \choose k}]{\prod_{m \in S_k} \, a_m} \, ,$$

while for $r \, \neq \, 0$ we have

$$M_{r,k} (a_j) \ := \ \left( \frac{1}{{n \choose k}} \sum_{m \in S_k} \, \prod \, a_m^r \right)^{1/r} \, ,$$

so we get $G \, = \, M_{0,1}(a_j)$, $A \, = \, M_{1}(a_j) \, = \, M_{1,1}(a_j)$, $A \, = \, M_{-1}(a_j) \, = \, M_{-1,1}(a_j).$

Example: $$M_{5,2} (a_1, a_2, a_3) \ = \ \left( \frac{1}{{3 \choose 2}} \left( a_1^5a_2^5 \, + \, a_1^5a_3^5 \, + \, a_2^5a_3^5 \right) \right)^{1/5}$$

The identity

$$M_{0,k}(a_j) \ = \ ( M_{0,1}(a_j))^k$$

for $k \, \in \, \{1, \ldots,n \}$ is then verified by using the appropriate identity for binomial coefficients. The salient point of considering these generalizations is then, that we can generalize the identity $G^2 \, = \, AH$ for two numbers to

$$( M_{0,1}(a_j))^n \ = \ M_{r,k}(a_j) \, \cdot \, M_{-r,n-k}(a_j) \, ,$$

where

$$( M_{0,1}(a_j))^{n} \ = \ M_{r,n/2}(a_j) \, \cdot \, M_{-r,n/2}(a_j) \, ,$$

valid for even $n$, is only the most symmetric generalization of $G^2 \, = \, AH.$

The identity

$$( M_{0,1}(a_j))^n \ = \ M_{r,k}(a_j) \, \cdot \, M_{-r,n-k}(a_j) \, ,$$

is proven simply by factoring $( M_{0,1}(a_j))^n$ from $M_{r,k}(a_j).$

This is the point, where we bring Newtons inequalities to bear, which read

$$\left( M_{1,1}(a_j) \right)^k \ \geq \ M_{1,k}(a_j)$$

for $k \, \in \, \{1, \ldots,n \}$ in our notation and equality is attained if and only if all $a_j$ are equal. Applying the special case $k \, = \, n-1$ to

$$M_{1,n-1}(a_j) \, \cdot \, M_{-1,1}(a_j)) \ = \ ( M_{0,1}(a_j))^n \, ,$$

immediately gives a generalization of your left inequality. For the right inequality we note

$$M_{-1,k} (a_j) \ = \ \frac{1}{ M_{1,k}(1/a_j)} \, ,$$

which follows directly from the definition for $k \, \in \, \{1, \ldots,n \}$. By Newtons inequality for $k \, = \, n-1$ applied to the set of $1/a_j$, $j \, \in \, \{1, \ldots,n \}$ we have

$$\frac{1}{M_{1,n-1}(1/a_j)} \ \geq \ \frac{1}{\left( M_{1,1}(1/a_j) \right)^{n-1}} \, ,$$

to which we apply

$$M_{-1,1} (a_j) \ = \ \frac{1}{ M_{1,1}(1/a_j)} \, ,$$

to obtain

$$M_{-1,n-1}(a_j) \ \geq \ \left( M_{-1,1}(a_j) \right)^{n-1} \, ,$$

from which the generalization of your right inequality follows by an application to

$$( M_{0,1}(a_j))^n \ = \ M_{1,1}(a_j) \, \cdot \, M_{-1,n-1}(a_j)) \, .$$

We obtain

$$M_{-1,k}(a_j) \ \geq \ \left( M_{-1,1}(a_j) \right)^{k}$$

for $k \, \in \, \{1, \ldots,n \}$ analoguesly to the special case $k \, = \, n-1.$

One can write down further generalizations of your inequalities by applying the other cases of Newtons inequalities to the equation

$$( M_{0,1}(a_j))^n \ = \ M_{\pm 1,k}(a_j) \, \cdot \, M_{\mp 1,n-k}(a_j)$$

for the corresponding indices. There should be simple generalizations to other values of $r$ as well.

This way to consider the power means is really quite natural and I am looking for a reference to the literature for it.

Maybe you can look into the Generalized mean function, you can derive all basic means (A, G, P, H) from it, and get and prove many interesting inequalities, including the one you stated there :)

https://en.wikipedia.org/wiki/Generalized_mean

• Thanks, but I checked the link and I couldn't really find the inequality that I referred to in my question. Could you point out where I can find it? – Matt L. Jun 3 '13 at 11:58