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...



Schur positivity on 2 letter alphabets implies Schur-positivity on n letters?
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 
$$P(x\_1,x\_2,x\_3)=s\_\emptyset(x\_1)(s\_3(x\_2,x\_3)+s\_{21}(x\_2,x\_3))+2s\_1(x\_1)s\_2(x\_2,x\_3)+2s\_2(x\_1)2s\_1(x\_2,x\_3)+s\_3(x\_1)s\_\emptyset(x\_2,x\_3).$$


