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I want to prove the claim that for all $n\geq 1,$ we have $${n\choose r} > {n\choose r+1}$$ if and only if $$n-1 \geq r > \frac{1}{2}(n-1).$$

Assume first that $${n\choose r} > {n\choose r+1}.$$ Then we notice that $n\choose r+1$ is defined if and only if $r + 1 \leq n$ which gives us $r\leq n-1.$ Further, from ${n\choose r} > {n\choose r+1}, $ we have \begin{align} \frac{n!}{(n-r)!r!} &> \frac{n!}{(n-r-1)!(r+1)!} \\ \frac{n!}{(n-r)(n-r-1)!r!} &> \frac{n!}{(n-r-1)!(r+1)r!} \\ \frac{1}{n-r} &> \frac{1}{r+1} \\ r+1 &> n - r \\ \end{align} which gives us the desired inequality. For the converse, assume $$n-1 \geq r > \frac{1}{2}(n-1).$$ Since $r \leq n-1 < n,$ both $n \choose r+1$ and $n \choose r$ are defined. From, $r > \frac{1}{2}(n-1),$ we have $r+1 > n-r$. Also, $n - r > 0$ so \begin{align} \frac{1}{n-r} &> \frac{1}{r+1} \\ \frac{n!}{(n-r)(n-r-1)!r!} &> \frac{n!}{(n-r-1)!(r+1)r!} \\ \frac{n!}{(n-r)!r!} &> \frac{n!}{(n-r-1)!(r+1)!} \\ \end{align} which, by definition gives us the inequality we needed.

Please see if my given proof is correct and suggest ways to improve the same.

Bill Dubuque
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Chirag Kar
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    Correct and similar to https://math.stackexchange.com/questions/857553 – Anne Bauval Dec 26 '22 at 21:48
  • Seems to me that $r+1\leq n$ should be an assumption to even talk about the statement (since you do not define binomial numbers otherwise). In such case you are in fact only proving $r > \frac{1}{2}(n-1)$ inequality, and your argument could be just equivalence instead of two implications (shorter). Also note that $\binom{n}{r}$ is often defined for $r>n$, it's just $0$ (follows directly from the combinatorial definition). But then $r=n$ satisfies your inequality and yet $r\not\leq n-1$, so again in that case $r \leq n-1$ would have to be assumption in the statement itself. – Sil Dec 26 '22 at 22:05
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    Posting can be shortened by asserting that $$\binom{n}{r+1} = \binom{n}{r} \times \frac{n-r}{r+1} ~: ~n \in \Bbb{Z^+}, ~r \in {0,1,2,\cdots,n-1}.$$ – user2661923 Dec 26 '22 at 22:14
  • You are showing that the sequence $\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}$ is unimodal. A nice presentation of this is given here. – Sammy Black Dec 26 '22 at 23:27

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