Recall the constructions $[n]_q=\frac{1-q^n}{1-q}, [n]!_q=[1]_q[2]_q\cdots[n]_q$ with $[0]!_q:=1$ and the $q$-binomials (Gaussian polynomials) $$\binom{n}k_q=\frac{[n]!_q}{[k]!_q[n-k]!_q}.$$ Given two polynomials $f(q)$ and $g(q)$, we write $f(q)\geq g(q)$ provided that $f(q)-g(q)$ is a polynomial having non-negative coefficients.
I would like to ask:
QUESTION. Suppose $0\leq k\leq a<b$ are integers. Is it true that $\binom{b+a}{b-k}_q\geq\binom{a+b}{a-k}_q$?
We know that there is a $q$-unimodality of the $q$-binomial coefficients. That is, $\binom{n}{k}_q - \binom{n}{k-1}_q$ has nonnegative coefficients for $k \leq n/2$. This was shown by Lynne M. Butler in A unimodality result in the enumeration of subgroups of a finite abelian group (in more generality than just $q$-binomials).
Now we just observe that $b-k$ is closer to the center peak at $(a+b)/2$ then $a-k$ is. So, $\binom{b+a}{b-k}_q - \binom{a+b}{a-k}$ has nonnegative coefficients as desired.
Just to add an alternating approach to that of Butler and Andrews, let's show that $\binom{n}k_q-\binom{n}{k-1}_q\geq0$, provided $2k\leq n$.
Let $n=\alpha k+d$ where $0\leq d<k$. Rewrite \begin{align*} \binom{n}k_q-\binom{n}{k-1}_q &=q^k\binom{n}{k-1}_q\frac{1-q^{(\alpha-2)k}}{1-q^k} +q^{(\alpha-1)k}\binom{n}{k-1}_q\frac{1-q^{d+1}}{1-q^k}. \end{align*} Observe $\frac{1-q^{(\alpha-2)k}}{1-q^k}$ is already a polynomial with non-negative coefficients. Furthermore, since $U(q):=\binom{n}{k-1}_q$ is unimodal, the coefficient of $q^j$ in $U(q)\cdot(1-q^{d+1})$ is non-negative as long as $2j\leq\deg(U)$. The same is true for $U(q)\frac{1-q^{d+1}}{1-q^k}$ as a formal power series. Since the polynomial $U(q)\frac{1-q^{d+1}}{1-q^k}$ is symmetric, having degree no greater than $U(q)$, all remaining coefficients of $U(q)\frac{1-q^{d+1}}{1-q^k}$ are non-negative.