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Define the function $D: \mathbb{R}_+^n \to \mathbb{R}$ by $$D(x_1, \dots, x_n) = \frac{1}{n} \sum_{i=1}^n x_i - \left( \prod_{i=1}^n x_i \right)^{1/n},$$ the difference between the arithmetic mean and geometric mean.

By the arithmetic-mean geometric-mean (AMGM) inequality, we know that $$0 \le D(x_1, \dots, x_n)$$ for all choices of points $x_1, \dots, x_n \ge 0$. We further know that $D(x_1, \dots, x_n) = 0$ if and only if $x_1 = \dots = x_n$, so that there is an upper bound $D(x_1, \dots, x_n) \le 0$ if and only if $x_1 = \dots = x_n$.

This leads me to wonder whether there is a (nontrivial) upper bound, say $B(x_1, \dots, x_n)$, for $D(x_1, \dots, x_n)$ in general which, for instance, satisfies that $B(x_1, \dots, x_n) = 0$ if and only if $x_1 = \dots = x_n$. Is there?

Note, I'm particularly interested in being able to quantify statements such as: if the points $x_1, \dots, x_n$ are "nearly equal", then the arithmetic mean and geometric mean will be "nearly equal".

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    $\begingroup$ Related: projecteuclid.org/download/pdf_1/euclid.rmjm/1181071887 $\endgroup$
    – Jack D'Aurizio
    Jan 13, 2018 at 16:36
  • $\begingroup$ Shouldn't $0$ be a lower bound for $D$? $\endgroup$
    – Anton Grudkin
    Jan 13, 2018 at 17:05
  • $\begingroup$ @AntonGrudkin It is indeed an upper bound when all of the input are equal. I'll edit to make this more clear. Sorry for the confusion! $\endgroup$
    – user163964
    Jan 13, 2018 at 20:45
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    $\begingroup$ See for example this answer for some choices of $\,B\,$. $\endgroup$
    – dxiv
    Jan 14, 2018 at 1:31

1 Answer 1

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As demonstrated in this paper here, J.M. Aldaz demonstrates that for the positive numbers $a_1, \ldots, a_n$ and the weights $w_1, \ldots w_n$ the inequality below holds: $$\sum_{k=1}^n w_k(a_k^{1/2}-\mu)^2\leq \sum_{k=1}^n w_k a_k-\prod_{k=1}^n a_k^{w_k}$$ where $$\mu=\sum_{k=1}^n w_ka_k^{1/2}\,.$$ This inequality captures the case of equality in the more usual AM-GM Inequality. This inequality is also nice in that both sides have the same degree of homogeneity.

The proof is quite elementary and based on the statistical fact that $E(X-\mu)^2=E(X^2)-(EX)^2$ for a random variable. But I'll let you read Aldaz's paper yourself.

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  • $\begingroup$ Thank you for sharing this! This provides a tighter lower bound than zero, like the paper mentioned by Jack D'Aurizio in the comments. However, my question is specifically asking for upper bounds. $\endgroup$
    – user163964
    Jan 17, 2018 at 17:37
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    $\begingroup$ @user163964 The ideas in Aldaz's paper allow you to conclude bounds such as $$D(x_1, \ldots, x_n)\leq \sum_{k=1}^n w_kx_k-\left(\sum_{k=1}^n w_k x_k^{-\epsilon}\right)^{-1/\epsilon}$$ for every positive $\epsilon$. Is that nice enough to be useful to you though? These bounds don't have the benefit of manifesting $L^2$ like Aldaz's main result. $\endgroup$
    – user123641
    Jan 17, 2018 at 18:36

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