Strengthening the Log-Concavity of Binomial Coefficients: $\binom{n}{k-1}\binom{n}{k+1} < \binom{n}{k}^2 - \binom{n}{k}$

Question

In the following question:

Log concavity of binomial coefficients: $ \binom{n}{k}^2 \geq \binom{n}{k-1}\binom{n}{k+1} $

It is proven via a combinatorial injective argument.

However, by noticing that the set of identical pairs of choices cannot be in the output space of the transformed pair of selections, we can strengthen the inequality to

$$\binom{n}{k-1}\binom{n}{k+1} \leqslant \binom{n}{k}^2 - \binom{n}{k}$$

From which the natural question follows: Is there a combinatorial argument that allows us to remove the equality case?

(This has been checked numerically and holds)

That is, how can we extend the argument even further to prove

$$\binom{n}{k-1}\binom{n}{k+1} < \binom{n}{k}^2 - \binom{n}{k}$$

Answer 1

Left Hand Side

Company $X$ wish to distribute $k-1$ candies and $k+1$ chocolates to $n$ kids. The number of possible way to achieve this is given by the LHS:

$$ \binom{n}{k-1}\binom{n}{k+1} $$

alternative expression

$m\in\left[0,k-1\right]$ kids receive both candies and chocolates, $k-m-1$ receive only candies, and $k-m+1$ receive only chocolates (so $2k-m$ different kids get something). If we sum over all possible $m$ we also get the total number of possible distribution:

$$ \binom{n}{k-1}\binom{n}{k+1} = \sum_{m=0}^{k-1}\binom{n}{2k-m}\binom{2k-m}{m}\binom{2k-2m}{k-m-1} $$

Right Hand Side

Another company, company $Y$ wish to distribute $k$ candies and $k$ chocolates to $n$ kids. However, company rule prohibit a distribution where $k$ kids get both candy and chocolate. The number of possible ways is given by the RHS:

$$ \binom{n}{k}\binom{n}{k}-\binom{n}{k} $$

alternative expression

$m\in\left[0,k-1\right]$ kids receive both candies and chocolates, $k-m$ receive only candies, and $k-m$ receive only chocolates (so $2k-m$ different kids get something). If we sum over all possible $m$ we also get the total number of possible distribution:

$$ \binom{n}{k}\binom{n}{k}-\binom{n}{k} = \sum_{m=0}^{k-1}\binom{n}{2k-m}\binom{2k-m}{m}\binom{2k-2m}{k-m} $$

Putting Everything Together

It's well known that $\binom{2k-2m}{k-m-1}<\binom{2k-2m}{k-m}$ so we have what we need:

$$ \binom{n}{k-1}\binom{n}{k+1} < \binom{n}{k}\binom{n}{k}-\binom{n}{k} $$

Answer 2

I will work in the context of the injection argument. Note that inequality only holds for when $n > k > 0$.

We see that the $\binom{[n]}{k-1} \binom{[n]}{k+1}$ sets correspond to the sets $(X, Y)$ in $\binom{[n]}{k}^2$ for which there exists some $i$ such that the number of elements up to $i$ in $X$ exceeds the number of elements up to $i$ in $Y$. We also see that the $\binom{n}{k}$ in your statement refers to the $(X, Y)$ in $\binom{[n]}{k}^2$ for which $X = Y$, so won't be mapped to by the injection.

But we also have the following: For some subset $A$ of $\{3, 4, 5, \dots, n\}$ of size $k-1$, consider the sets $X=\{2\} \cup X$ and $Y = \{1\} \cup X$. Then again $(X, Y)$ is not mapped to, so thus inequality cannot occur.

Answer 3

What you want to prove is false in the case where $k=0$ or $k=n$:$$\binom{n}{-1}\binom{n}{1}=\binom{n}0^2-\binom n0.$$

However, in all other cases, what you want to prove is true. In fact, we can prove something much stronger. You wanted to prove that $$\binom{n}{k}^2-\binom{n}{k-1}\binom{n}{k+1}\stackrel{?}>\binom nk.$$ Starting on the LHS, we obtain $$ \begin{align} \binom{n}{k}^2-\binom{n}{k-1}\binom{n}{k+1} &= \binom{n}{k}^2- \left[\frac{k}{n-k+1}\binom nk\right] \left[\frac{n-k}{k+1}\cdot \binom n k\right] \\&= \binom{n}{k}^2\left[1-\frac{k(n-k)}{(n-k+1)(k+1)}\right] \\&= \frac{n+1}{(k+1)(n-k+1)}\binom nk^2 \\&= \color{blue}{\frac1{k+1}\binom{n+1}{k}}\binom nk. \end{align} $$ Note that the blue factor out front, $\color{blue}{\frac1{k+1}\binom{n+1}{k}}$, is always strictly greater than $1$. To see this, write this as $$ \color{blue}{\frac1{k+1}\binom{n+1}{k}}=\color{blue}{\frac1{n+2}\binom{n+2}{k+1}}, $$ and note that, since $1\le k\le n-1$, we always have $\binom{n+2}{k+1}\ge \binom{n+2}2=(n+2)\cdot \frac{n+1}2>(n+2)$.