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.