Inequality involving binomial coefficient

Question

I am reading a proof in which it is claimed that for all $2 \leq k \leq n-2$, the following inequality holds $$k(n-k) < {n \choose k} - 1.$$

Why is this true? I've tried to show this by induction on $n$ and I've also tried to find different objects to use in a counting argument, but I've had no success so far.

I would also like to learn general techniques for deriving such inequalities. I feel a little clueless about how to get started on this one.

To provide more information, this appears in the second proof here that for any $2 \leq k \leq n-2$, there exists a $k$-covector on $\mathbb{R}^n$ which is not a blade, i.e., not decomposable into a wedge product of $k$ covectors.

Edit: I originally made a mistake and had written a nonstrict inequality. I have modified the inequality to be strict.

Answer 1

Hints:

• By AM-GM $\;k(n-k) \le \frac{n^2}{4}\,$.

• The binomial coefficients $\binom{n}{k}$ for fixed $n$ and variable $k$ are monotonically increasing for $k=1,2,\dots,\lfloor n/2 \rfloor$ then monotonically decreasing from $k=\lfloor (n+1)/2 \rfloor\,\dots,n\,$ (see here for example). Therefore $m = \min \left\{\binom{n}{k} \mid 2 \le k \le n-2\right\}=\binom{n}{2}=\frac{n(n-1)}{2}\,$.

• $2 \leq k \leq n-2$ implies $n \ge 4\,$, and $\frac{n^2}{4} \lt \frac{n(n-1)}{2} - 1$ for $n \ge 4\,$.

Piecing together:

$$k(n-k) \le \frac{n^2}{4} \lt \frac{n(n-1)}{2} -1 = m -1 \le \binom{n}{k} -1$$

Answer 2

Not a very rigorous answer, but:

wlog assume $n-k > k$.

We can rewrite the inequality as :

$n(n-1)(n-2)\cdots (n-k+1) - 1 \geq k(n-k)k!$

There are k+2 "things" being multiplied together on the LHS, and k+2 "things" on the RHS.

Clearly $n > n-k, \\ n-1 > k, \\ n-2 \geq k, n-3 \geq k-1 \\ n-4 \geq k-2 \\ \vdots \\ n-k+1 \geq 1$\

therefore the LHS is strictly greater than the RHS, and since all are integers (assuming?), we have the extra "-1", changing it to a soft (nonstrict) inequality.