1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.