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.