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 $n$.) 



Lower-bound for $\Pr[X \geq m]$ subject to $E[X]>m$ where $X$ is a binomial random variable
$\newcommand{\al}{\alpha}
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\De}{\Delta}$
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. 


