The AM-GM inequality is a known one, it states that for any list of n numbers, the following is true:

$$ \frac{x_1 + x_2 + \text{...} + x_n}{n} \ge \sqrt[n]{x_1 \cdot x_2 \cdot \text{...} \cdot x_n }$$

But there are other important statistic means in math, such as the harmonic mean:

$$ H =\left(\frac{\sum_{k=1}^n x_k^{-1}}{n}\right)^{-1}$$

Or the root mean square:

$$ X_\text{rms} = \sqrt{\frac{1}{n} (x_1^2 + x_2 ^ 2 + ... + x_n^2)} $$

The AM-GM can be extended with these means. More specifically, given any list of numbers:

let A be the arithmetic mean, GM be the geometric mean

let H be the harmonic mean and $X_\text{rms}$ bet the root mean square

The following is true:

$$ X_\text{rms} \ge A \ge GM \ge H $$

All of these inequalities can be proven using mathematical induction, and it is a well-known proof. But I want to consider a different approach:

$ \text{For any list of n numbers, let's define the function }$

$$ S(p) = \left( \frac{x_1^p + x_2^p + \text{...} + x_n^p}{n} \right) ^ \frac{1}{p} $$

$ \text{Then} $ $ H = S(-1) $, $ A = S(1) $, $ X_\text{rms} = S(2)$

$ \text{Notice that}$ $GM = \lim_{p \to 0} S(p) $

So we can say for sure that $S(2) \ge S(1) \ge S(p \to 0) \ge S(-1) $. We then make the hypothesis that $ S(x) \text{is monotonically increasing on the set of real numbers }$

My question is how to prove this hypothesis? Ideally, your answer would also contain a proof of $\lim_{p \to 0} S(p) = GM$ as I figured this one out on pure intuition.

P.S I am in the last grade of school right now, so no high-level math (Higher than basics of calculus or complex numbers theory) would be prefered. Still, though, any help would be very much appreciated!

  • $\begingroup$ You should at least learn asymptotic expansion (see links from my profile), so that you can immediately confirm systematically yet rigorously your guess that the power mean tends to the geometric mean as the power tends to zero. $\endgroup$
    – user21820
    Sep 26, 2019 at 7:50

1 Answer 1


This is shown in many places.

There is a proof using Jensen's inequality in https://en.wikipedia.org/wiki/Generalized_mean

One book I like is in chapter 8 of

THE CAUCHY–SCHWARZ MASTER CLASS An Introduction to the Art of Mathematical Inequalities by J. MICHAEL STEELE

It's available in both soft-cover and digital.



and the classic Inequalities by G. H. Hardy, J. E. Littlewood, G. Pólya

  • $\begingroup$ Are there any free online resources where I can see this proof? I think it would be strange to buy a mathematics book for only one proof... $\endgroup$
    – Philippe
    Sep 25, 2019 at 17:13
  • $\begingroup$ I did a search for "generalized mean" and found the reference I added to my answer. $\endgroup$ Sep 25, 2019 at 18:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .