Pick integers $b\geq a \geq 0$ and $k\geq j\geq 0$. It is not super-difficult to prove the inequality $$ \binom{kb}{ka}^j \geq \binom{jb}{ja}^k. $$ This is actually quite a nice inequality that was used in a paper of mine joint with N. Amini (with a slight generalization), but to prove it required several intermediate steps.

Now consider the following $q$-analogue. Is it true that $$ \binom{kb}{ka}_q^j \geq q^{a(b-a)j(k-j)k/2} \binom{jb}{ja}_q^k $$ where the inequality now means that the coefficient of $q^i$ in the left hand side, is always greater-than-or-equal to the coefficient of $q^i$ in the right hand side?

I have verified my suspicions in the cases $b,k\leq 6$.

EDIT: Inspired by Fedor's comment, it seems helpful to rewrite the inequality as follows, where now $c,d\geq 0$, and still $k\geq j$. $$ q^{k cd \binom{j}{2}}\binom{kc+kd}{kc}_q^j \geq q^{j cd \binom{k}{2}}\binom{jc+jd}{jc}_q^k. $$ Both sides can now be interpreted as sums over lattice paths from $(0,0)$ to $jk(c,d)$ using only up and right steps. The $q$-weight is simply the area between the path and the $x$-axis. However, the left hand side requires that the path passes through the points $k(c,d),2k(c,d),3k(c,d),\dotsc,jk(c,d)$, while the right hand side requires that the paths passes through $j(c,d), 2j(c,d),3j(c,d),\dotsc,jk(c,d)$.

There is now some intuition, as we have fewer requirements on the paths in the left hand side, so this count should be larger. In fact, if $k$ is a multiple of $j$, the inequality is obviously true, as the left hand side count a strict super-set of paths. But some details are of course missing.

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    $\begingroup$ The non-$q$ version may be phrased as follows: take a $k\times j$ chessboard and put $b$ coins at each square. We choose $kja$ coins, LHS counts the number of ways to choose them so that each column contains exactly $ka$ chosen coins, RHS --- so that each row contains exactly $ja$ chosen coins. Is there a combinatorial/probabilistic interpretation of the $q$-version? $\endgroup$ May 3, 2019 at 10:34
  • $\begingroup$ Well, the coefficient of $q^i$ in $\binom{kb}{ka}_q$ is the number of binary strings of length $kb$ with $ka$ ones, that has major index (or number of inversions) equal to $i$. One can also interpret this as lattice paths in a $k(b-a)$-rectangle between two fixed opposite corners, and then count how many of these has $i$ squares below the path. $\endgroup$ May 3, 2019 at 10:39


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