Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a followup to this question Is this AM/GM refinement correct or not?. In it, as joriki helpfully pointed out, the correct form of a generalization of AM/GM proved by Dragomir is as follows. Let $p_i=1$ and $x_i\in[a,b]$ then the following holds: $$ \exp\left(\frac{1}{2b^2n^2}\sum\limits_{i<j} (x_i-x_j)^2\right)\le\frac{A_n(x)}{G_n(x)} $$ Please consult the original question for the full form of the inequality.

Now I am asking the following question. What is an effective lower bound on the $A_n/G_n$ ratio for $x_i\in[0,1]$ if we know that $\min|x_i-x_j|\ge p$, i.e. the numbers are not close to each other.

Case n=2

Let $f_2(x,y)=(x+y)/(2 \sqrt{x y})$ and assume $|x-y|\ge p$. Using Dragomir's inequality we get the following lower bound: $f(x,y)>\exp(p^2/8)$.

If we however examine the function $f_2(x,x-p)$ on the interval $[0,1]$ we see that it is decreasing in $x$ (and increasing in $p$) and it's minimum (for fixed $p$) is attained at $x=1$. Hence we arrive at the following lower bound $f_2(x,y)\ge f_2[1,1-p]$, or in other words: $$\frac{x+y}{2 \sqrt{x y}}\ge \frac{2-p}{2 \sqrt{1-p}}>\exp(p^2/8)$$

So this is an improvement on the general inequality.

Case n=3

Let $f_3(x,y,z)=(x+y+z)/\sqrt[3]{xyz}$ where $x$,$y$, and $z$ lie in [0,1] and the minimal distance between any two of them is $p<1/2$. According to the general inequality, we have $f_3(x,y,z)>\exp(p^2/3)$.

Let's examine the function $f_3(x,x-p,x-q)$ where $x\in[0,1]$ and $x>\max(p,q)$. It's derivative is surprisingly simple:$$\frac{df_3(x,x-p,x-q)}{dx}=\frac{x \left(-2 p^2+2 p q-2 q^2\right)+p^2 q+p q^2}{9 x (x-p)(x-q) \sqrt[3]{x (p-x) (q-x)}}$$

Simple calculations show that the minimum is achieved when $x=1$ and this minimum is: $$f_3(1,1-p,1-q) = \frac{3-p-q}{3 \sqrt[3]{(1-p) (1-q)}} $$ Taking again the derivative it is obvious that given a fixed $q$, the minimum is attained when $p=q/2$, or in other words $f_3(1,1-p,1-q)\ge f_3(1,1-p,1-2p)$ for $q>p$.

We have thus proved that: $$\frac{x+y+z}{3 \sqrt[3]{x y z}}\ge f_3(1,1-p,1-2p)=\frac{1-p}{\sqrt[3]{2 p^2-3 p+1}}$$

So again we have a better bound then the one offered by Dragomir's inequality, though I can't prove that it is better.

The questions.

  1. How can one prove that $\frac{1-p}{\sqrt[3]{2 p^2-3 p+1}}>\exp(p^2/3)$? I have verified this only numerically.
  2. How can those results be generalized for $n>3$ and hence improve the bound $A_n/G_n\ge \exp(\frac{1}{24} \left(n^2-1\right) p^2)$ which follows from Dragomir's inequality?
  3. Are those inequalities known already?


The two cases I discussed can be straight-forward generalized to the following hypothesis for the general case. Let $q\ge n-1$, $x_i\in[0,1]$, and $|x_i-x_j|>=1/q$. Then the following inequality holds $$\frac{A_n(x)}{G_n(x)}>\left(\frac{1-n}{2}+q\right) \left((-1)^n \left(-q\right)_n\right)^{-1/n}$$where $\left(-q\right)_n$ is the Pochhammer symbol. Note that for $n<4$ this is equivalent to what's proved above.

share|cite|improve this question
You should try to see if you can use majorization: – kjetil b halvorsen Sep 27 '12 at 7:36
You are very right! It is obvious from what I wrote in my answer below that the $A_n/G_n$ ratio is Schur-convex. This is another way to arrive at the $x_i=x-(i-1)p$ vector. – ivan Sep 27 '12 at 10:57
up vote 1 down vote accepted

It turns out I can almost prove this myself.

Please note that questions 1 and 3 and the finish of this proof are still missing.

Lets denote the arithmetic/geometric mean of $x_n,x_{n-1},...x_i$ with $A_i$, resp. $G_i$ The trick is to notice that:$$\frac{d}{dx_1}\frac{A_n}{G_n}=\frac{x_1-A_n}{x_1 G_n}$$ and hence the minimum of $A_n/G_n$ for fixed $x_i,i>1$ is achieved when $x_1=A_{n}$ and this is equivalent to $x_1=A_{n-1}$!

Let's evaluate $A_n/G_n$ at $x_1=A_{n-1}$. We get:$$\left .\frac{A_n}{G_n}\right\vert_{x_1=A_{n-1}} = \left(\frac{A_{n-1}}{G_{n-1}}\right)^{\frac{n-1}{n}}$$

Continue in the same way to take derivatives to see that minimum is achieved for $x_i=A_{n-i}$. When we are down to two variables we need $x_n=x_{n-1}$ for minimum. Of course it follows from here that the minimum for our ratio is 1 and is achieved when all variables are equal. However we can't have that because $|x_i-x_j|>=p$.Backtracking our steps now shows that the minimum is achieved for $x_i-x_{i+1}=p$ for each $i<n$. At this $x_i$ (letting $x_i=x-(i-1)p$) the value of the ratio is:$$\frac{A_n}{G_n}=\left(\frac{1-n}{2}+\frac{x}{p}\right) \left((-1)^n \left(-\frac{x}{p}\right)_n\right)^{-1/n}$$ Technically we need to prove that this function is decreasing in $x$ for $x\in [n/p,1]$. This I don't know how to do. The derivative (up to a positive factor) is :$$((n-1) p-2 x) \psi ^{0}\left(-\frac{x}{p}\right)+(-np+p+2 x) \psi ^{0}\left(n-\frac{x}{p}\right)+2 n p$$ where $\psi^{0}$ is the digamma function.

It suffices to prove of course that this derivative is negative in our interval, but I don't see how to do this except perhaps by using a fine enough approximation with a power series.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.