We say two polynomials satisfy $P(x)\geq Q(x)$ iff $P(x)-Q(x)$ has non-negative coefficients. Recall $(n)_q!=\prod_{j=1}^n(1-q^j)$ and the Gaussian polynomials $\binom{n}k_q=\frac{(n)_q!}{(k)_q!(n-k)_q!}$.
A very interesting conjecture (only partially solved) by F. Bergeron, The q-Foulkes Conjecture, Talk delivered at Bowdoin College, slides found here, states that if $0<a\leq b\leq c\leq d$ and $ad=bc$ are integers then $$\binom{b+c}b_q\geq\binom{a+d}a_q.$$ After some experimentation, I arrived at:
QUESTION. Suppose $0<a_1\leq b_1\leq c_1\leq d_1, 0<a_2\leq b_2\leq c_2\leq d_2, a_1d_1=b_1c_1, a_2d_2=b_2c_2$ and $a_1+b_2+c_2+d_1=a_2+b_1+c_1+d_2$, then is it true that the coefficients of $$\binom{a_1+d_1}{a_1}_q\binom{b_2+c_2}{b_2}_q-\binom{b_1+c_1}{b_1}_q\binom{a_2+d_2}{a_2}_q$$ alternate in signs exactly twice as: $+ - +$? I accept any proof even assuming Bergeron's conjecture.
REMARK. Let $0<a_1\leq b_1\leq c_1\leq d_1, 0<a_2\leq b_2\leq c_2\leq d_2, a_1d_1=b_1c_1, a_2d_2=b_2c_2$. Assuming Bergeron's conjecture, it is easy to prove that $$\binom{b_1+c_1}{b_1}_q\binom{b_2+c_2}{b_2}_q\geq\binom{a_1+d_1}{a_1}_q\binom{a_2+d_2}{a_2}_q.$$
