Question: Can Chebyshev's sum inequality be used to prove the generalized mean inequality, or at least a portion of it?
If we let $$P(r) = \sqrt[r]{\frac{x_1^r + x_2^r + \cdots +x_n^r}{n}}$$, we have that $P(r)$ is increasing, which is also known as the generalized mean inequality/power mean inequality. The most intuitive proof for me is via Jensen's, but I came up with the following and was intrigued by it's promise.
Chebyshev's sum inequality states that if $a_1 \geq a_2 \geq \cdots \geq a_n$ and $b_1 \geq b_2 \geq \cdots \geq b_n$, then we have $$ \frac{1}{n} \sum_{k=1}^{n} a_k b_k \geq \left ( \frac{1}{n} \sum_{k = 1}^{n} a_k \right ) \left ( \frac{1}{n} \sum_{k = 1}^{n} b_k \right ) $$
Indeed, we can use this to prove the QM-AM inequality (the 2-nd power mean is $\geq$ the 1st power mean) fairly easily, using Chebyshev's inequality on $x_1, x_2, \cdots, x_n$, and $x_1, x_2, \cdots, x_n$ which we can assume is increasing.
We have $$ \frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n} \geq \left ( \frac{x_1+x_2+\cdots+x_n}{n} \right)^2 $$
directly from Chebyshev's, which shows $P(2) \geq P(1)$.
A three-dimensional Chebyshev would not be very hard to prove using a three-dimensional Rearrangement inequality, but from the intuition of Rearrangement, it seems clear that if $a_1 \leq a_2 \leq \cdots a_n$, $b_1 \leq b_2 \leq \cdots b_n$, and $c_1 \leq c_2 \leq \cdots c_n$, we have for any permutations $\sigma$ and $\pi$ of the set $\{1, 2, \cdots, n\}$ we have $$ a_1b_1c_1 + a_2b_2c_2 + \cdots + a_nb_nc_n \geq a_1b_{\sigma(1)}c_{\pi(1)} + a_2 b_{\sigma(2)}c_{\pi(2)} + \cdots + a_n b_{\sigma(n)} c_{\pi(n)} $$ from which a three variable Chebyshev (I think) should follow by summing rearrangement inequalities formed by cyclic permutations of the set $1, 2, \cdots, n$. (I'll post details of this after I work all of them out.)
That is,
$$ \frac{1}{n} \sum_{k=1}^{n} a_k b_kc_k \geq \left ( \frac{1}{n} \sum_{k = 1}^{n} a_k \right ) \left ( \frac{1}{n} \sum_{k = 1}^{n} b_k \right ) \left ( \frac{1}{n} \sum_{k=1}^{n} c_k \right ) $$
And from this we obtain $P(3) \geq P(1)$. Assuming this multivariable generalization is valid, it is not hard to show $P(k) \geq P(1)$ for all (integer) $k$. Some sleight of hand might be possible to show it for all real $k$ as well.
Is it possible to show that $P(3) \geq P(2)$, though? Or in general, $P(m) \geq P(n)$ for integers $m, n$? I would be quite satisfied if it worked for just integers, and would be intrigued if anyone has a full prove for reals, which I suspect wouldn't be too hard from the integral form of Chebyshev's sum inequality.
Update 1: I have conjectured the following more general version of Chebyshev's Sum Inequality, which would prove the Power Mean Inequality directly.
Let $\mathbf{P} = (p_1, p_2, \cdots, p_j)$ and $\mathbf{Q} = (q_1, q_2, \cdots, q_k)$ be partitions of a positive integer $n$. If $\mathbf{P}$ majorizes $\mathbf{Q}$, then for increasing sequences of length $n$: $a_{1m}, a_{2m}, \cdots, a_{nm}$, we have $$ \left ( \frac{1}{n} \sum_{k = 1}^{n} \prod_{i=1}^{p_1} a_{ik} \right ) \left ( \frac{1}{n} \prod_{i=p_1+1}^{p_2} a_{ik} \right ) \cdots \left ( \frac{1}{n} \sum_{k = 1}^{n} \prod_{i=p_{j-1}+1}^{p_j} \right ) \geq \left ( \frac{1}{n} \sum_{k = 1}^{n} \prod_{i=1}^{q_1} a_{ik} \right ) \left (\frac{1}{n} \prod_{i=q_1+1}^{q_2} a_{ik} \right )\cdots \left ( \frac{1}{n} \sum_{k = 1}^{n} \prod_{i=q_{k-1} + 1}^{q_k} \right ) $$
or something like this, similar to Muirhead's inequality. I'll write more details as they come to me.
Is it possible to prove the latter inequality with Generalized-Chebyshev? Applying it with $r$ sets of $z_1^{1/s}, z_2^{1/s}, \cdots, z_n^{1/s}$ yields
$$ \frac{1}{n}\sum z_i^{r/s} \geq \left ( \frac{1}{n} \sum z_i^{1/s} \right )^r $$
but I'm not sure if I can bound the RHS with $\frac{1}{n} \sum z_i$ if I take the $rs$-th root.
– WeierstrassSauce Jul 18 '13 at 15:28