TITLE: Inequality on permutation polytope
QUESTION [6 upvotes]: Let $a = (a_1, \cdots, a_n), b = (b_1, \cdots, b_n), c = (c_1, \cdots, c_n) \in \mathbb{R}^n$ with $a_1 \geq \cdots \geq a_n, b_1 \geq \cdots \geq b_n, 0 < c_1 \leq \cdots \leq c_n$.
In addition, assume that $\sum_{i=1}^k b_i \leq \sum_{i=1}^k a_i$ for all $k \in \{1, \cdots, n-1\}$ and $\sum_{i=1}^n b_i = \sum_{i=1}^n a_i$.
Let $A :=  \{(a_{\sigma(1)}, \cdots, a_{\sigma(n)}) \,|\, \sigma \in S_n\}$ and $K_A$ be the unique minimal convex set containing $A$.
For $b \in K_A$, does $$\sum_{i=1}^n c_i b_i \geq \sum_{i=1}^n c_i a_i$$ hold?

REPLY [3 votes]: Write $c_k=p_1+\cdots+p_k$ with $p_k=c_k-c_{k-1}\ge0$. Then
$$\sum_ic_ib_i-\sum_ic_ia_i=\sum_kp_k(s_k(b)-s_k(a))$$
with $s_k(a)=a_k+\cdots+a_n$. By assumption, $s_k(b)-s_k(a)\ge0$ and you are gone.

This proof is just Abel's resummation, that is discrete integration by parts.