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In some paper the authors make use of the following inequality without further explanation: Let $xinmathbbR^n$ with $x_1lecdotsle x_n$ and $alphain[0,1]^n$ with $sum_i=1^n alpha_i=Nin1,2,ldots,n$. Then $$sum_i=1^nalpha_i x_igesum_i=1^N x_i.$$

While I already have found a (quite lengthy) bare-hands-proof, I wonder if this inequality is just (some variant of) some commonly known inequality that I am just unaware of. Any hints?

(In a way this is a rephrasing of Iosif's answer, but from a different perspective. My main intention is to point out the connection to matroids.)

The inequality follows from:

Fact. The polytope $K = Biglalpha in [0,1]^n: sum_i = 1^nalpha_i = NBigr$ is the convex hull of indicator vectors of subsets of $[n] = 1, ldots, n$ of size $N$.

Assuming this fact, and because any linear function over a polytope achieves its minimum at a vertex, we have that $sum_i = 1^nalpha_i x_i ge sum_i in S x_i$ for some set $S$ of size $N$. The right hand side is then clearly minimized for $S = 1, ldots, N$ if $x_1 le x_2 le ldots le x_n$.

To show the Fact, we need to show that every extreme point $alpha$ of $K$ is an indicator vector of a set of size $N$. This is clearly true if $alpha in 0,1^n$, so assume, towards contradiction, that for some $i$ we have $0 < alpha_i < 1$. Then there must be at least one other $j neq i$ such that $0 < alpha_j < 1$, and we can write $alpha$ as a convex combination of two vectors in $K$, one in which we add a tiny $h$ to $alpha_i$ and subtract $h$ from $alpha_j$, and one in which we reverse the signs. This means that $alpha$ is not an extreme point, a contradiction. You can see that this argument is essentially the same as Iosif's.

Another way to state the Fact is that $K$ is the base polytope of the uniform matroid over $[n]$ of rank $N$. A vast generalization is the characterization of the facets of the base polytope of any matroid, due to Edmonds: see, e.g., Section 10.7 in these notes.

Use Abel transform: denote $x_i=y_1+y_2+dots+y_i$, then $$sum alpha_i x_i=sum y_i (alpha_i+alpha_i+1+dots+alpha_n). $$
We have $alpha_i+alpha_i+1+dots+alpha_n=N-(alpha_1+dots+alpha_i-1)geqslant N-i+1$ for $i=1,dots,N$. Therefore
$$
sum alpha_i x_igeqslant Ny_1+(N-1)y_2+dots+y_N=x_1+dots+x_N.
$$

$newcommandalalpha$
The following, rather intuitive proof is done by induction on $n$. The case $n=1$ is trivial. Suppose now that $nge2$. By continuity, without loss of generality $x_1<dots<x_n$. The minimum of $sum_1^nal_i x_i$ over all $al=(al_1,dots,al_n)$ as in the OP is attained. Let $al=(al_1,dots,al_n)$ be a point of such an attainment.

To obtain a contradiction, suppose that $al_1<1$. Then the condition $sum_1^nal_i=Nge1$ implies that $al_j>0$ for some $jin2,dots,n$. Replacing $al_1$ and $al_j$ respectively by $al_1+h$ and $al_j-h$ for a small enough $h>0$, we get a smaller value of $sum_1^nal_i x_i$ (because $x_1<x_j$). This contradicts the assumption that $al$ is a point of minimum of $sum_1^nal_i x_i$.

So, $al_1=1$, and your inequality reduces to $sum_2^nal_i x_igesum_2^N x_i$ given that $al_iin[0,1]$ for all $i$ and $sum_2^nal_i=N-1$, and the latter inequality is true by induction.