Symmetric ratio of binomial probabilities around the midpoint $\frac{m+1}{2}$ greater $\frac{p}{1-p}$ for $p>0.5$ I am trying to show the following: 
$\frac{P_r\left(\tfrac{m+1}{2}\leq X \leq \frac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \frac{m-1}{2}\right)}\geq \frac{p}{1-p}$,
with $n>m\geq \frac{n+1}{2}$,  $1>p>0.5$ and  $n,m$ are both odd Natural Numbers with $n\geq3$. 
The probabilities I am considering come from a binomial distribution, so that I get
$\frac{\displaystyle \sum_{i=\frac{m+1}{2}}^{\frac{n-1}{2}}\binom{m}{i}p^i (1-p)^{m-i}}{\displaystyle \sum_{i=\frac{2m-n+1}{2}}^{\frac{m-1}{2}} \binom{m}{i}p^i(1-p)^{m-i}}\geq\frac{p}{1-p}$
It is worth noting that the concerned intervals lie symmetrically around $\frac{m}{2}$ on the line $0,\ldots,m$: 
$m-\frac{m+1}{2}=\frac{2m-n+1}{2}=\frac{2m-n+1}{2}-0$ and 
$\frac{m+1}{2}-\frac{m}{2}=\frac{1}{2}=\frac{m}{2}-\frac{m-1}{2}$.
Now a as $p>0.5$, it seems intuitive that the probability above the midpoint is higher than the one below it. But I just don't know how to show that it igretaer than $\frac {p}{1-p}$. All my other results indicate that it must be true, but I am unable to show it directly. 
Since I have absolutely no idea how to tackle this problem, I would appreciate any hints, ideas or help. 
Thanks in advance. 
Update:
I have played around a bit with the above expression and reformulated it. Maybe this helps? 
$P_r(\tfrac{m+1}{2}\leq X \leq \tfrac{n-1}{2})=P_r(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2})-P_r(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2})$
Therefore, 
$\frac{P_r\left(\tfrac{m+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}= \frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)-P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}=\frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}-1 $
Thus 
$\frac{P_r\left(\tfrac{m+1}{2}\leq X \leq \frac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \frac{m-1}{2}\right)}\geq \frac{p}{1-p} \iff \frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}-1 \geq \frac{p}{1-p}$ 
$\iff \frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)} \geq \frac{1}{1-p} \iff (1-p) \geq \frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}$
$\iff
1-\frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)} \geq p
\iff \frac{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)-P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{m-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)} \geq p$
$\iff \frac{P_r\left(\tfrac{m+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)} \geq p$
From here on, I don't know how to show that:
$\frac{P_r\left(\tfrac{m+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}{P_r\left(\tfrac{2m-n+1}{2}\leq X \leq \tfrac{n-1}{2}\right)}\geq p$ 
 A: Note that
\begin{align}
\sum_{i=(2m-n+1)/2}^{(m-1)/2} \binom{m}{i}p^i(1-p)^{m-i} &=~\sum_{i=(m+1)/2}^{(n-1)/2} \binom{m}{m-i}p^{m-i}(1-p)^i \\
&=~\sum_{i=(m+1)/2}^{(n-1)/2} \binom{m}{i}p^{m-i}(1-p)^i
\end{align}
Therefore,
\begin{align}
&\frac{\Pr\left(\tfrac{m+1}{2}\leq X \leq \frac{n-1}{2}\right)}{\Pr\left(\tfrac{2m-n+1}{2}\leq X \leq \frac{m-1}{2}\right)} =~\frac{\sum_{i=(m+1)/2}^{(n-1)/2} \binom{m}{i}p^{i}(1-p)^{m-i}}{\sum_{i=(m+1)/2}^{(n-1)/2} \binom{m}{i}p^{m-i}(1-p)^i} \\
\geq~&\min\left\{\frac{\binom{m}{(m+1)/2}p^{(m+1)/2}(1-p)^{(m-1)/2}}{\binom{m}{(m+1)/2}p^{(m-1)/2}(1-p)^{(m+1)/2}},\ \cdots,\ \frac{\binom{m}{(n-1)/2}p^{(n-1)/2}(1-p)^{(2m-n+1)/2}}{\binom{m}{(n-1)/2}p^{(2m-n+1)/2}(1-p)^{(n-1)/2}}\right\} \\
=~&\min\left\{\frac{p^{(m+1)/2}(1-p)^{(m-1)/2}}{p^{(m-1)/2}(1-p)^{(m+1)/2}},\ \cdots,\ \frac{p^{(n-1)/2}(1-p)^{(2m-n+1)/2}}{p^{(2m-n+1)/2}(1-p)^{(n-1)/2}}\right\} \\
=~&\min\left\{\left(\frac{p}{1-p}\right)^{\frac{m+1}{2} -\frac{m-1}{2}},\ \cdots,\ \left(\frac{p}{1-p}\right)^{\frac{n-1}{2}-\frac{2m-n+1}{2}}\right\} \\
=~&\min\left\{\left(\frac{p}{1-p}\right)^1,\ \cdots,\ \left(\frac{p}{1-p}\right)^{n-m-1}\right\} \\
=~& \frac{p}{1-p}
\end{align}
where the first inequality is because $\frac{a_1 + \cdots + a_n}{b_1 + \cdots + b_n} \geq \min\{\frac{a_1}{b_1}, \cdots, \frac{a_n}{b_n}\}$ for positive $a_i$s and $b_i$s.
