Let $x_1\leq x_2\leq \ldots\leq x_k\leq y\leq x_{k+1}\leq\ldots \leq x_n$. Let $p>1$. Does the following inequality is true? $$\sum_{i=1}^k\left(\frac{1}{n+1}((y-x_i)^p+\sum_{j=i+1}^n(x_j-x_i)^p)\right)^{1/p}+\left(\frac{1}{n+1}\sum_{j=k+1}^n(x_j-y)^p\right)^{1/p}$$ $$+\sum_{i=k+1}^n\left(\frac{1}{n+1}\sum_{j=i+1}^n(x_j-x_i)^p\right)^{1/p}\geq\sum_{i=1}^n\left(\frac{1}{n}\sum_{j=i+1}^n(x_j-x_i)^p\right)^{1/p}$$

Mainly I would like to ask for suggestions for showing it (I think it is true). It is connected with the generalized mean (it came from the following problem: consider the generalized mean of $\max\{x_j-x_i,0\}$ for a given $i$, and $x_1\leq\ldots\leq x_n$. Then take a sum over $i$. Will it increase if we will add $y\in(x_1,\ldots,x_n)$ to the group?). I tried to write it as a sum of $\ell_p$ norms and I tried to use concavity of $x^{1/p}$ and superadditivity of $x^p$ but it did not work. Maybe I'm missing something.

  • $\begingroup$ Working on this problem I came to the inequality, which if occurs then I am able to show the above inequality: $$n\sum_{i=1}^n \left|x_i-\frac 1n \sum_{k=1}^n x_k\right|^p\geq \sum_{i=1}^n\sum_{j=i+1}^n(x_j-x_i)^p$$ but still I don't see how to show this one. $\endgroup$ – FF2 May 20 '15 at 7:24

Too long for a comment.

I have tested both of your inequalities for $y=0$ and the following three suspicious cases:

1) $x_1=-1$, $x_2=\dots=x_{n-1}=0$, $x_n=1$.

2) $x_1=\dots=x_{n/2}=-1$, $x_{n/2+1}=\dots=x_{n}=1$.

3) $x_1=\dots=x_{n-1}=-1$, $x_n=n-1$.

In all of tested cases the computational evidence suggests that the long inequality (from the question) holds for all $n$ and $p$, whereas the asymptotic suggests that short inequality (from the question) fails for some large $n$ and $p$. More detailed:

1) The long inequality becomes


The short inequality becomes

$$2n\ge 2^p+2n-4.$$

2) The long inequality becomes

$$\frac n2\left(\frac{n\cdot 2^{p-1}+1}{n+1}\right)^{1/p}+\left(\frac{n/2}{n+1}\right)^{1/p}\ge \frac n2 \left(\frac{n\cdot 2^{p-1}}{n}\right)^{1/p}.$$

The short inequality becomes

$$n^2\ge\frac{n^2\cdot 2^p}{4}.$$

3) The long inequality becomes

$$(n-1)\left(\frac{n^p+1}{n+1}\right)^{1/p}+\left(\frac{(n-1)^p}{n+1}\right)^{1/p}\ge (n-1)\left(\frac{n^p}{n}\right)^{1/p}.$$

The short inequality becomes

$$n((n-1)+ (n-1)^p)\ge (n-1)n^p.$$


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