Let's start from a classical inequality:
If $0\le a_1\le\cdots\le a_k$ and $0\le b_1\le\cdots\le b_k$ then $(a_1+\cdots+a_k)(b_1+\cdots+b_k)\le k(a_1b_1+\cdots+a_k b_k)$.
It can be written also in the form of averages: $Av(\{a_i\})Av(\{b_i\})\le Av(\{a_ib_i\})$ (expressing convexity and many other properties)
I need the following generalization for finite partially ordered sets. Let $S$ be such a set, consider non-decreasing functions, i.e. $f:S\rightarrow\Bbb{R}_{\ge0}$, satisfying: if $a\ge b$ then $f(a)\ge f(b)$.
I need: $\underline{\text{ if $f,g$ are non-decreasing functions then } Av_S(f)Av_S(g)\le Av_S(fg).}$
If $S$ is totally ordered, then one gets the classical version.
The inequality does not hold for arbitrary partial ordered sets (with obvious counterexamples). I guess a necessary condition is that $S$ has minimal and maximal elements. Even this is not enough (with obvious counterexamples).
In my particular case $S$ is the set of lattice points on a simplex, i.e.: $S_{n,r}:=\{(k_1,\dots,k_r)|\ k_1+\cdots+k_r=n,\ k_1,\dots,k_r\ge0\}$. (One can think about this as the set of monomials in r variables of total degree n.) The order is induced by recursive application of the rule $x^2_i\ge x_ix_j$. (So, e.g. $x^n_i\ge x^{n-1}_ix_j\ge x^{n-2}_ix_jx_k\ge\cdots$.) And the considered functions are symmetric (i.e. invariant w.r.t. to the permutation group $\Xi_r$, that acts on $S_{n,r}$.)
Alternatively, one can consider the quotient $S_{n,r}/\Xi_r$. (This set is partially ordered, with minimal and maximal elements.)
Probably in this particular case the inequality is well known? I guess, a necessary condition on a partially ordered set to satisfy such an inequality (for any non-decreasing functions) is that $S$ is "ordered enough". Can this be made precise? Are there some sufficient conditions known?
For bookkeeping: in a very particular case (here) we proved this bound by terrible brute force.
upd: we have proved this inequality (for $S_{n,r}$) arXiv:1412.8200.