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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Edit: I originally made a mistake and had written a nonstrict inequality.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.

Matthew Kvalheim
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2 Answers2

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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:

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

dxiv
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  • Thanks for the help. I was hoping there would be a simpler answer, because I don't know how I would come up with this on my own, but at least the steps make sense to me. – Matthew Kvalheim Jul 17 '17 at 04:59
  • Would you mind elaborating a bit on your intuition or how you came up with this solution? Is it just from practice? E.g., I wouldn't have known to expect that AM-GM is "sharp enough" that it wouldn't spoil the rest of the estimate. – Matthew Kvalheim Jul 17 '17 at 05:02
  • @MatthewKvalheim The inequality itself is rather weak. What the above proves is in fact the stronger statement $k(n-k) \lt \binom{n}{k'} - 1$ for any $2 \le k, k' \le n-2,$. As far as intuition, I knew that the binomial coefficients go "up and down" from $k=1$ to $k=n$ and was just checking on how that relates to the LHS, when I realized that the stronger statement does in fact hold. Given how loose the proposed inequality is (and, maybe, the combinatorics tag as well) I would not be surprised in the least if a more elegant one-line combinatoric proof existed. – dxiv Jul 17 '17 at 05:08
  • Thanks for elaborating. Just one more thing: can you offer any intuition about using AM-GM specifically, or did you just try it arbitrarily and saw that it was sufficient? – Matthew Kvalheim Jul 17 '17 at 05:27
  • @MatthewKvalheim AM-GM was just the most succinct way to justify it. You could as well look at $,f(k)=k(n-k) = -k^2 + n,k,$ as a quadratic function in $k,$, and note that it has a maximum (since the leading coefficient is negative) at $k = n/2,$, which maximum is $f(n/2) = n^2 / 4,$. – dxiv Jul 17 '17 at 05:32
  • Ok, thanks very much for your help. – Matthew Kvalheim Jul 17 '17 at 05:37
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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.

CRX111011
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