A GENERAL inequality for a bi-modal hypergeometric distribution Say $X$ has a hypergeometric distribution with parameters $m$, $n$ and $k$, with $k\leq n<\frac12m$.
I know that $X$ has a dual mode if and only if $d=\frac{(k+1)(n+1)}{m+2}$ is integer. In that case $P(X=d)=P(X=d-1)$ equals the maximum probability.
See my previous question. I got a great answer proving $P(X=d+1) > P(X=d-2)$. That got me wondering: can we make a more general statement? More specifically (for natural $c \leq d-2$):
$P(X=d+c) > P(X=d-1-c)$
This is true for $c = 1$, but also in many cases when $c \geq 2$. I have not found any counterexamples yet. Can this be proven? Or where to start?

You can turn the answer from the other question into an inductive proof for this question*.
$$\tfrac{P(X=d+c+1)}{P(X=d+c)}-\tfrac{P(X=d-c-2)}{P(X=d-c-1)} = \tfrac{(k-d-c)(n-d-c)}{(d+1+c) (m-k-n+d+1+c)} -\tfrac{ (d-c-1) (m-k-n+d-c-1)}{(k-d+c+2)(n-d+c+2)} \\= \tfrac{(k-d-c)(n-d-c)(k-d+c+2)(n-d+c+2)-(d-c-1) (m-k-n+d-c-1)(d+1+c) (m-k-n+d+1+c)}{(d+1+c) (m-k-n+d+1+c)(k-d+c+2)(n-d+c+2)}$$
again the denominator is positive, and we only need to show that the numerator is positive.
We can do the same steps, substituting $d=(k+1)(n+1)/(m+2)$ gives for the numerator:
$$(c+1)^2(m-2k)(m-2n)$$
which is positive when both $k< \frac{1}{2}m$ and $n < \frac{1}{2}m$.

Some other interesting points


*

*For $c = 0$ you get the previous answer. 

*For $c=-1$ you get $\frac{P(X=d)}{P(X=d-1)}-\frac{P(X=d-1)}{P(X=d)} = 0$, which is true by the assumption $P(X=d) = P(X=d-1)$.

*Also for $n=\frac{1}{2}m$ you get that the term $(m-2n)$ equals zero and you get the symmetry $P(X=d+c) = P(X=d-c-1)$ 

*If $\tfrac{P(X=d+c+1)}{P(X=d+c)}-\tfrac{P(X=d-c-2)}{P(X=d-c-1)}> 0$ and $$P(X=d+c) \geq P(X=d-1-c)$$ then $P(X=d+(c+1)) > P(X=d-1-(c+1))$ 
