generalized mean inequality related problem 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. 
 A: 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 
$$n((n-1)+ (n-1)^p)\ge (n-1)n^p.$$
