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.

rearrangement inequality, which gives the absolute maximum arrangement. The statement made naively analogously for the poset case is false. (b) Whether the statement holds is very sensitive to the shape of the poset. The poset with $\{0,1\}$ and $\{a,b\}$ such that $0 < a < 1$ and $0 < b < 1$ satisfies your inequality. But if you make the middle step three elements big instead of two it fails. $\endgroup$ – Willie Wong Dec 6 '13 at 13:09