I am currently working on the following problem:
Given two sets $A = \{a^1,a^2, \ldots ,a^{N}\}$ and $B = \{b^1,b^2, \ldots ,b^{M}\}$ with $a^k,b^k \in \mathbb{R}^d, d \geq 2$ and the knowledge that $\forall a \in A \; \exists b \in B: a \leq_d b$ and $\forall b \in B \; \exists a \in A: a \leq_d b$ and moreover $a_i \not \leq_d a_j \; \forall i \neq j$ and $b_i \not \leq_d b_j \; \forall i \neq j$ I am trying to show that:
$\frac{1}{N} \sum\limits_{k=1}^N a^k \leq_d \frac{1}{M} \sum\limits_{k=1}^M b^k$
I started to suppose that there is a component $1 \leq i \leq d$ such that $M \sum\limits_{k=1}^N a^k_i > N \sum\limits_{k=1}^M b^k_i$. But this just lead me to nowhere. Of course for every $a_i^k$ on the left side I will find some $b \in B$ with $b_i \geq a_i^k$ and for every $b_i^k$ on the right side I would find some $a \in A$ with $b_i^k \geq a_i$. But that didn't take me to the contradiction I was looking for.
I am currently stuck here and hope to get some further with your help.
Edit: $\leq_d$ is meant to be the natural order on $\mathbb{R}^d$, so $a \leq_d b :\Leftrightarrow a_i \leq b_i \;\text{ for all } 1 \leq i \leq d$.
The inequality which your are trying to show needs many conditions to be satisfied, so it looks generically wrong. As a concrete counterexample we can pick $d=3$, $N=3$, $M=2$, $A=\{(10,8,0), (8,10,0),(0,0,6)\}$, and $B=\{(10,10,0), (0,0,6)\}$. Then
$$\frac{1}{N} \sum\limits_{k=1}^N a^k =(6,6,2) \not\leq_d (5,5,3)= \frac{1}{M} \sum\limits_{k=1}^M b^k.$$