Showing monotonicity for ratio of binomial pmf and tail cdf I'm interested in showing for $X\sim\text{Bin}(n,p)$, $p\in(0,1)$ that when $x\geq np$,
$$
\frac{P(X=x)}{P(X\geq x)}\leq \frac{P(X=x+1)}{P(X\geq x+1)}
$$
I've verified using numerical simulations, but can't seem to get the algebra to work out. 
Below I plot the ratio $P(X=x)/P(X\geq x)$ for values of $x\geq \lfloor np\rfloor$. We can see that the ratio is increasing with $x$. Furthermore, I plot the difference between the ratio evaluated at $x+1$ (the right hand side above) and the ratio evaluated at $x$ (the left hand side above) for varying $x$. The difference is also positive, though the difference increases for $x$ close to $np$ and then decreases thereafter to zero.
Numerical Simulations
n=500
p=0.3
x=seq(floor(n*p),n)
ratio=dbinom(x,n,p)/pbinom(x-1,n,p,lower.tail = FALSE)
diff=dbinom(x,n,p)/pbinom(x-1,n,p,lower.tail = FALSE)-dbinom(x-1,n,p)/pbinom(x-2,n,p,lower.tail = FALSE)

plot(x,ratio)
plot(x,diff)



 A: We know that
$$
\frac{P(X=x+1)}{P(X=x)}=\frac{n-x}{x+1}\frac{p}{1-p}
$$
Therefore, we will show that 
$$
\frac{P(X\geq x+1)}{P(X\geq x)}\leq \frac{n-x}{x+1}\frac{p}{1-p}
$$
Multiplying the left hand side by $(1-p)/p$ and expanding out the numerator and denominator on the left hand side we have
\begin{align}
\frac{1-p}{p}\frac{P(X\geq x+1)}{P(X\geq x)}&=\frac{1-p}{p}\frac{\sum_{k\geq x+1}\dbinom{n}{k}p^k(1-p)^{n-k}}{\sum_{k\geq x}\dbinom{n}{k}p^k(1-p)^{n-k}}\\
&=\frac{\binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+\binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+\binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+\cdots}{\binom{n}{x}p^{x+1}(1-p)^{n-x-1}+\binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+\binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+\cdots}\tag{1}
\end{align}
Note that the numerator has $n-x$ terms while the denominator has $n-x+1$ terms. Now we examine ratios of successive binomial coefficients. Note that 
$$
\frac{\binom{n}{x+1}}{\binom{n}{x}}=\frac{n-x}{x+1}
$$
and for $k\geq 1$
$$
\frac{\binom{n}{x+1+k}}{\binom{n}{x+k}}=\frac{n-x-k}{x+1+k}\leq \frac{n-x}{x+1}
$$
Combining the last two results, we may write for $k\geq 0$
$$
\binom{n}{x+1+k}\leq \frac{n-x}{x+1}\binom{n}{x+k}
$$ 
Substituting this inequality for each binomial coefficient in the numerator of (1), we have
\begin{align}
\frac{1-p}{p}\frac{P(X\geq x+1)}{P(X\geq x)}&=\frac{\binom{n}{x+1}p^{x+1}(1-p)^{n-x-1}+\binom{n}{x+2}p^{x+2}(1-p)^{n-x-2}+\binom{n}{x+3}p^{x+3}(1-p)^{n-x-3}+\cdots}{\binom{n}{x}p^{x+1}(1-p)^{n-x-1}+\binom{n}{x+1}p^{x+2}(1-p)^{n-x-2}+\binom{n}{x+2}p^{x+3}(1-p)^{n-x-3}+\cdots}\\
&\leq \frac{n-x}{x+1}\frac{\sum_{k=x+1}^n \binom{n}{k-1}p^k(1-p)^{n-k} }{\left[\sum_{k=x+1}^n \binom{n}{k-1}p^k(1-p)^{n-k}\right] +p^{n+1}(1-p)^{-1} }\\
&\leq \frac{n-x}{x+1}
\end{align}
as desired. 
