Set of solutions for a binomial inequality I bumped into the following inequality:
$${a-b\choose c}{a\choose c}^{-1} \le \exp\left(-\frac{bc}{a}\right)$$
Playing with it a little bit, trying to bound it asymptotically for large $a$'s, using Stirling's approximation, I ended up with nothing. Finally I decided to put some numbers and check it, and figured out it is wrong. Moreover: it looked like it is always wrong.
Can you prove this inequality?
Edit: since it wasn't clear, I'll add that $a,b,c$ are all positive integers.
Edit: I also forgot to add the assumption $a>b>c$.

I will now prove the inequality, using probability techniques. I wonder if this is the easiest way to show it...
Theorem 2.10 in $\color{blue}{[1]}$ shows Chernoff-type bounds may be applied to Hypergeometric distributed random variables. In particular, if $X$ is a random variable, $X\sim\text{Hypergeometric}(N,K,n)$, $\mu=\mathbb{E}(X)=nKN^{-1}$, then
\begin{equation*}
  \mathbb{P}(X\le \mu-a) \le \exp\left(-\mu\varphi\left(\frac{-a}{\mu}\right)\right)
\end{equation*}
for $a\ge 0$ and $\varphi=(1+x)\ln(1+x)-x$ for $x\ge -1$ (and $\varphi(x)=\infty$ for $x<-1$).
Now, let $X\sim\text{Hypergeometric}(a,b,c)$. The LHS of our inequality is then simply $\mathbb{P}(X=0)$. While $\varphi$ is not well defined for $x=-1$, we may continuously extend it to have $\varphi(-1)=1$. Hence
  \begin{equation*}
    \frac{{a-b\choose c}}{{a\choose c}}
    = \mathbb{P}(X\le \mu-\mu) \le \exp\left(-\mu\varphi\left(-\frac{\mu}{\mu}\right)\right)
    = \exp(-\mu) = \exp\left(-\frac{bc}{a}\right).
  \end{equation*}
References


*

*$\color{blue}{[1]}$ S. Janson, T. Łuczak and A. Ruciński, Random Graphs, New York, Wiley (2000)

