Set of solutions for a binomial inequality

Question

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 $a>b>c$.

Answer 1

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)

Answer 2

Actually, I don't know if you want LHS $\ge $ RHS or LHS $\le $ RHS.

But here is an example: If you take $a=b+1=c+2$ --- e.g. $a=5,b=4,c=3$, then LHS $=0$ and the RHS is positive. So LHS $<$ RHS.

I think that given your restrictions $a>b>c$, it can never hold that LHS $\ge $ RHS.

Answer 3

Without further restrictions on $a,b,c$ neither inequality can be correct: Let $l(a,b,c)$ the LHS and $r(a,b,c)$ the RHS, then $$l(3,2,1)=1/3 < r(3,2,1) = \exp(-2/3) \approx 0.513417$$ but $$l(3,2,-1/2)=35/24 \approx 1.458333 > r(3,2,-1/2) = \exp(1/3) \approx 1.395612$$