# Proving the Power Mean Inequality using Chebyshev's sum inequality

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.

• What stops you from applying the same mathod? Use change of variable $x_i^{r/s}\to z_i$. (I didn't use $n/m$ for the exponent to avoid confusion with the number of variables in the sum.) – S.B. Jul 18 '13 at 4:24
• Thanks for your suggestion! Maybe I'm misinterpreting your comment, but using the same method on $s$ sets of $x_1^{r/s}, x_2^{r/s}, \cdots, x_n^{r/s}$ yields $$\frac{x_1^r + x_2^r + \cdots + x_n^r}{n} \geq \left ( \frac{x_1^{r/s} + x_2^{r/s} + \cdots + x_n^{r/s}}{n} \right )^s$$, and I'm not sure how that helps. – WeierstrassSauce Jul 18 '13 at 4:50
• Sorry, I think I should've said use change of variable $x_i^s\to z_i$. So we want to rewrite $$\left(\frac{1}{n}\sum x_i^r\right)^s\geq\left(\frac{1}{n}\sum x_i^s\right)^r.$$ Now take $x_i^s$ on the RHS it becomes simply $z_i$. On the LHS each $x_i^r$ becomes $z_i^{r/s}$. Take the $r$-th root from both sides and you'll get $$\left(\frac{1}{n}\sum z_i^{r/s}\right)^{s/r}\geq \frac{1}{n}\sum z_i.$$ – S.B. Jul 18 '13 at 13:17
• Ah, good point! 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
• I don't know why you changed the inequality again. I made it in the form that you preferred (set $t=r/s$ in what I arrived at). The exponent on the RHS are all ones as you liked. So you only need to prove what you showed for positive integers in the question, for all real numbers. Basically, it suffices to show that $P(t)\geq P(1)$ for all $t\geq 1$. – S.B. Jul 18 '13 at 15:46