Lower-bound for $\Pr[X \geq m]$ subject to $E[X]>m$ where $X$ is a binomial random variable Given an integer number $m>0$ and a real number $\alpha\in [1, 2]$,  I am interested in finding a lower-bound for $\Pr[X\geq m]$ subject to $X \sim \text{Binomial}(n, m\alpha/n)$.
For large values of $m$, I use Chernoff bound to get my desired bound. However, for smaller values of $m$, I need a tighter bound. Based on my observations (using Mathematica) for any $m > 0$ and $\alpha\in [1, 2]$, $\Pr[X\geq m]$ is minimized when $n\rightarrow \infty$, in which case I can use the relation between Binomial and Poisson distribution to compute $\Pr[X\geq m]$ numerically. 
So, here is my question: is it correct that for any $\alpha \in [1, 2]$ and $m>0$, $\Pr[X\geq m]$ is minimized subjecto to $X \sim \text{Binomial}(n, m\alpha/n)$ when $n\rightarrow \infty$?
An observation that might be helpful is that this claim is not correct for  $\Pr[X > m]$. However, even this function seems to be minimized either when $n\rightarrow \infty$ or $n = \lceil  m\alpha \rceil$(which is the minimum value possible for $n$.) 
 A: $\newcommand{\al}{\alpha}
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\De}{\Delta}$
The answer to your question is yes, and the restriction $\al\le2$ is not even needed. 
Indeed, let $S_n:=X\sim\text{Binomial}(n,m\al/n)$, where $m$ is any natural number, $\al$ is any real number $\ge1$, and $n$ is any natural number $\ge m\al$, so that $\al\in[1,n/m]$. It suffices to show that $P_n:=P_{n,\al}:=P_{n,\al,m}:=P(S_n\ge m)$ is nonincreasing in natural $n\ge m\al$. By Lemma 1 at the end of this answer, 
\begin{equation*}
 P_n=\frac{n!}{(m-1)!(n-m)!}\,J_n, \tag{1}
\end{equation*}
where 
\begin{equation*}
J_n:=J_{n,\al}:=J_{n,\al,m}:= \int_{1-m\al/n}^1 t^{n-m}(1-t)^{m-1}\,dt. 
\end{equation*}
It remains to show that $P_{n+1}/P_n\le1$, which can be rewritten as 
\begin{equation*}
 \De_{n,\al}:=(n+1)J_{n+1,\al}-(n-m+1)J_{n,\al}\overset{\text(?)}\le0. \tag{2}
\end{equation*}
For $\al\in(1,n/m)$,
\begin{align*}
 &\pd{\De_{n,\al}}\al \, \al (m\al)^{-m} (n+1)^{m-1} \left(1-\frac{m\al}{n}\right)^{m-n}\\ 
& = \De_{n,\al;1}:=
\left(1-\frac{m\al}{n}\right)^{m-n} \left(1-\frac{m\al}{n+1}\right)^{n-m+1}-\frac{n-m+1}{n}\,\left(\frac{n+1}{n}\right)^{m-1}.  
\end{align*} 
Further, for $\al\in(1,n/m)$, 
\begin{align*}
 &\pd{\De_{n,\al;1}}\al
 =\frac{(\al-1) m^2  n^{n-m} (n+1-m\al)^{n-m} }{(n+1)^{n-m+1} (n-m\al)^{n-m+1}} >0. 
\end{align*}
So, $\De_{n,\al;1}$ may change its sign only from $-$ to $+$ as $\al$ increases from $1$ to $n/m$. So, $\pd{\De_{n,\al}}\al$ has the same sign pattern. So, to verify the inequality in (2), it suffices to show that $\De_{n,0}\le0$ and $\De_{n,n/m}\le0$. 
Since $J_{n,0}=0$ for all $n$, we obviously have $\De_{n,0}=0\le0$. Modulo a proof of Lemma 1, it remains to note that 
$$\De_{n,n/m}=(n+1)J_{n+1,n/m}-(n-m+1)J_{n,n/m}
\le(n+1)J_{n+1,(n+1)/m}-(n-m+1)J_{n,n/m}
=1\Big/\binom n{m-1}-1\Big/\binom n{m-1}=0.$$


Lemma 1: If $Y\sim\text{Binomial}(n,p)$, then 
  \begin{equation*}
 g(p):=P(Y\ge m)=h(p):=\frac{n!}{(m-1)!(n-m)!}\,\int_{1-p}^1 t^{n-m}(1-t)^{m-1}\,dt 
\end{equation*}
  for all $p\in[0,1]$ and all natural $m$. 

Proof: We have $g(p)=\sum_{j=m}^n \binom nj p^jq^{n-j}$, with $q:=1-p$. So, for $p\in[0,1]$,
\begin{align*}
 g'(p)&=\sum_{j=m}^n \binom nj jp^{j-1}q^{n-j}
 -\sum_{j=m}^n \binom nj (n-j)p^jq^{n-j-1} \\
 &=n\sum_{j=m}^n \binom{n-1}{j-1} p^{j-1}q^{n-j}
 -n\sum_{j=m}^{n-1} \binom{n-1}j p^jq^{n-j-1} \\
 &=n\sum_{i=m-1}^{n-1} \binom{n-1}i p^iq^{n-1-i}
 -n\sum_{i=m}^{n-1} \binom{n-1}i p^iq^{n-1-i} \\
 &=n\binom{n-1}{m-1} p^{m-1}q^{n-m}=h'(p). 
\end{align*}
Also, $g(0)=0=h(0)$. Now Lemma 1 immediately follows. 
