Proof of the inequality $n(n-1)\dots (n-k+1)>(\frac{n}{2})^k$

Question

I am a little confused about this claim. This is part of a proof in Rudin's analysis book. He uses this claim as a fact without further proofs.

Claim If $n>2k>0$ and $n,k\in \mathbb{N}$, then $n(n-1)\dots (n-k+1)>(\frac{n}{2})^k$.

First of all, I failed to convince myself that this statement is true. By playing with algebra a little bit, this statement still does not seem obvious to me. To prove it, the induction theorem comes to me naturally. However, with the constriction that $n>2k$, I have to start from $n=2k+1$ as a base case. As a result, the base case even cannot be verified. Any hint will be appreciated.

Answer 1

$0<k<n/2$

$n-k+1>n-n/2+1=n/2+1>n/2$

$n(n-1)\dots (n-k+1)>(n-k+1)^k>(\frac{n}{2})^k$

Answer 2

Just to see why it's true, it's often useful to try it with some concrete numbers. Let's say $n=8$, $k = 3$. Then

$$8 \cdot 7 \cdot 6 \ge 4^3 = 4 \cdot 4 \cdot 4$$

Answer 3

Alternative Proof:

Assume that $$\frac{n}{n-m} \geq 2 \quad (0 \leq m \leq k)$$ $$\implies n \geq 2n - 2m$$ $$\implies n \leq 2m,$$ which contradiction shows that $$ \frac{n}{n-m} \lt 2,$$ for all $m$.

Next, note that there are $k$ terms in the product $$n(n-1) \cdots (n-k+1),$$ and so $$\frac{n^k}{n(n-1) \cdots (n-k+1)}= \frac{n}{n} \frac{n}{n-1} \cdots \frac{n}{n-k+1} < 2^k,$$ and a simple rearrangement gives the desired result.

$\square$