Proof about AM-GM inequality generalized

Question

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.

Answer 1

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

Answer 2

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