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Note: I'm not sure this type of questions are welcome on the site. In case tell me.

Let's define the $p$ mean as

$$M_p(x_1, \dots, x_n) = \sqrt[p] { \frac 1n \sum_{i = 1}^n x_i^p}$$

for $x_1, \dots, x_n > 0$.

Your goal is to prove that $$\dots \le M_{-2} \le M_{-1} = \mathcal{H} \le M_0 = \mathcal{G} \le M_1 = \mathcal{A} \le M_2 = \mathcal{Q} \le M_3 \dots$$

($M_0$ should be interpreted as the geometric mean, that is $M_0 = \sqrt[n]{\prod_{i = 1}^n x_i}$. The notation is justified if we consider $p \to 0$ )

Ideally I would like to have as many proof as possible of the above, the more elegant the better. Extra points for not using any "nuclear bomb" in proving the result!

P.S. In case it is not clear, $\mathcal{H}, \mathcal{G}, \mathcal{A}, \mathcal{Q}$ are respectively the harmonic, geometric, arithmetic and quadratic means.

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How are you defining $M_0$? – Mathmo123 Jul 10 '14 at 22:13
@Mathmo123 you are right! I edited :) – Ant Jul 10 '14 at 22:17
Are the $x_i$'s positive? Also, have you read this Wikipedia article? – JimmyK4542 Jul 10 '14 at 22:19
@JimmyK4542 I am not sure it needs to be so. I can see some troubles arising with the geometric mean though, so if you want to assume that, it is fine! :) – Ant Jul 10 '14 at 22:23
@Ant: consider $\{1/2,1/2,-1\}$. The arithmetic mean is $0$ but the harmonic mean is $1$. – robjohn Jul 10 '14 at 22:39
up vote 4 down vote accepted

For $p\le q$, $p\ne0$, and $q\gt0$, $x^{q/p}$ is convex on $x\gt0$. Therefore, Jensen's Inequality says that $$ \left(\frac1n\sum_{k=1}^nx_k^p\right)^{1/p}\le\left(\frac1n\sum_{k=1}^nx_k^q\right)^{1/q}\tag{1} $$ For $p\le q$ and $q\lt0$, $(1)$ follows by applying $(1)$ to $1/x_k$ with $-q\le-p$ and $-q\gt0$, which would yield the reverse inequality, but then taking the reciprocal of both sides reverses the inequality a second time.

Furthermore, $$ \begin{align} \lim_{p\to0}\left(\frac1n\sum_{k=1}^nx_k^p\right)^{1/p} &=\lim_{p\to0}\left(\frac1n\sum_{k=1}^ne^{p\log(x_k)}\right)^{1/p}\\ &=\lim_{p\to0}\left(\frac1n\sum_{k=1}^n\left[1+p\log(x_k)+O(p^2)\right]\right)^{1/p}\\ &=\lim_{p\to0}\left(1+O(p^2)+\frac pn\sum_{k=1}^n\log(x_k)\right)^{1/p}\\ &=\lim_{p\to0}e^{\frac1n\sum\limits_{k=1}^n\log(x_k)+O(p)}\\ &=\left(\prod_{k=1}^nx_k\right)^{1/n}\tag{2} \end{align} $$

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thanks! I wait before I accept it in case someone comes up with something different, even though as others have said the proof using convexity is basically the only one – Ant Jul 11 '14 at 7:30

The magic word is convexity, and there is really not much more to it.

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