So I need to prove something in the context of the binomial distribution. This particular bit is algebra-heavy, so I thought I'd post the question here. Basically, I can't seem to figure out what sort of manipulation took place that allowed the inequality in the first line to result in the inequality in the second line:
UPDATE: the specific context is to prove that
$m\le (n+1)p$
Given that $P(X=m) \ge P(X=m-1)$ and that $X$~$B(n,p)$.
The inequalities above are part of this solution:
This generally does not hold. (The reason you can't figure it out is because it's wrong!) I was playing around with it on desmos https://www.desmos.com/calculator/bwrji9t6ud and you'll see that it doesn't hold for $p$ too small. I can prove that the second inequality follows if we know that $p > 0.5$.
Firstly, by definition, you have \begin{align*}&{n\choose m} =\frac{n!}{m!(n-m)!},\\&{n\choose m-1} =\frac{n!}{(m-1)!(n-(m-1))!} = \frac{n!}{(m-1)!(n-m+1)!}\end{align*} Secondly, consider $p^m(1-p)^{n-m}$ and $p^{m-1}(1-p)^{n-m+1}$. From the context of these inequalities, $p$ is positive and nonzero, so $p^{m-1}$ is also positive and nonzero. This means we can divide both sides by $p^{m-1}$ without affecting the inequality. Upon division, you get $p(1-p)^{n-m}$ and $(1-p)^{n-m+1}$ respectively.
Let's focus on $p(1-p)^{n-m}$. I will show that it is less than or equal to $p^{n-m}$. We know that $n > m$ and $0 < p < 1$. (If $p=0$ or $p=1$, the inequality is trivial.) I will denote $n-m = x$, so we know that x is a non negative integer. We want to show that $p^x \geq p(1-p)^x$.
First consider the case where $x=0$. This holds trivially, as $p^0 = 1 > p = p(1-p)^0$. Now we can consider the case where $x ≥ 1$. Note that $p(1-p)^x = (p^{1/x}(1-p))^x$. Thus it suffices to show that $p ≥ p^{1/x}(1-p)$. Because $p$ is strictly greater than $0$ and less than $1$, $p^{1/x}$ gets larger as $1/x$ gets smaller. If we can show that the inequality holds for the smallest possible value of $1/x$, then we have shown that it holds for all values of $x$. Obviously there is no "smallest" value for $1/x$, it is just the limit as $x$ tends to infinity.
This case is also trivial given my assumption of $p > 0.5$. $\lim_{x\rightarrow\infty}p^{1/x} = p^0 = 1$. This leads to $p^{1/x}(1-p) = (1-p)$. Thus it is clear that if $p>0.5$, $p ≥ (1-p) ≥ p^{1/x}(1-p)$. (You can also see that $p >0.5$ is a necessity. if $p<0.5$, you can just find a large enough $x$ to reverse the inequality.)
This covers every case, so we end up with the inequality chain: \begin{align*} {n\choose m} p^m (1-p)^{n-m} &≥ {n\choose m-1} p^{m-1}(1-p)^{n-m+1}\\ \Rightarrow \frac{n!}{m!(n-m)!} p^m (1-p)^{n-m} &≥ \frac{n!}{(m-1)!(n-m+1)!} p^{m-1}(1-p)^{n-m+1}\\ \Rightarrow \frac{n!}{m!(n-m)!} p (1-p)^{n-m} &≥ \frac{n!}{(m-1)!(n-m+1)!}(1-p)^{n-m+1}\\ \mathrm{For}\ p > 0.5, \Rightarrow \frac{n!}{m!(n-m)!} p^{n-m} &≥ \frac{n!}{(m-1)!(n-m+1)!}(1-p)^{n-m+1} \end{align*}