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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.

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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.

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