TITLE: Schur positivity on 2 letter alphabets implies Schur-positivity on n letters?
QUESTION [7 upvotes]: Suppose we have a symmetric polynomial $P$ in $n$ variables.
We can partition this alphabet into sets with one or two letters, e.g. ${ {x_1}, {x_2, x_3}}. 
We can thus see $P$ as an element in $Q[x_1][x_2,x_3]$,
and expand it in the Schur basis in each sub-set of variables.
E.g, $P=5 s_{2}(x_1)s_{32}(x_2,x_3)$ or similar.
Suppose now that this expansion always have non-negative coefficients, for every choice of partition of letters into 1-or-2-element subsets.
Can we conclude that $P$ itself is Schur-positive in $n$ variables?
Note that the Littlewood-Richardson rule tells us that the converse is true.
If not, is there some $k = k(n)$ such that Schur-positivity on subsets of size $\leq k$ implies Schur-positivity in $n$ variables?
The intuition behind why 2-element sets might be enough is as follows: 
If we want to show that a sum over some combinatorial objects is Schur-positive, it suffices to create a map to SSYTs, or equivalently, to reading-words which are Knuth-equivalent to the SSYTs.
To completely describe a reading-word, it suffices to know how many times $i$ appear before $j$, for every pair $i$, $j$.
So, assume that I have a set of objects in bijection with SSYTs of shape $\lambda$. By expanding the sum over these it in different choices of 2-variable subsets, we can almost figure out this data above.
But I cannot make this formal, at least without using type A crystals or similar...

REPLY [3 votes]: My intuition is that if $P$ is in three variables then your requirements only force $P$ to be unimodal in each pair of variables, while being Schur positive is much more restrictive. For example, take 
$$P(x_1,x_2,x_3)=s_3(x_1,x_2,x_3)+s_{21}(x_1,x_2,x_3)-s_{111}(x_1,x_2,x_3).$$
Equivalently,
$$P(x_1,x_2,x_3)=m_3(x_1,x_2,x_3)+2m_{21}(x_1,x_2,x_3)+2m_{111}(x_1,x_2,x_3).$$
Now, partition the variables as $\{\{x_1\},\{x_2,x_3\}\}$, this partition is essentially unique. And then we have 
$$P(x_1,x_2,x_3)=s_\emptyset(x_1)(s_3(x_2,x_3)+s_{21}(x_2,x_3))+2s_1(x_1)s_2(x_2,x_3)+2s_2(x_1)2s_1(x_2,x_3)+s_3(x_1)s_\emptyset(x_2,x_3).$$