# Q-analogue of an inequality

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.

• 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? – Fedor Petrov May 3 at 10:34
• 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. – Per Alexandersson May 3 at 10:39