Recently I asked the following identity on Math.StackExchange knowing it had several proofs:
$$ \binom{n}{k}^2 \geq \binom{n}{k-1}\binom{n}{k+1} $$
See here. They quickly give the one-liner:
In your initial inequality divide the left hand side by the right hand side and simplify.
This is kind of a proof. It works on math.SE since many people know combinatorics there.
Probably they meant:
Using the identity $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ and cross multiplying, the identity is equivalent to: $$ k!^2 (n-k)!^2 \leq (k-1)!(k+1)! (n-k-1)! (n-k+1)! $$
Let's try proving: $k!^2 \geq (k-1)!(k+1)!$ Dividing both sides by $k!^2$ we get $1 \leq \frac{k+1}{k-1}$ which is true.
A similar proof works for the factors of $(n-k)!$
Overall, people had no idea where I was coming from. I had been reading a Richard Stanley article about log-concave sequence of numbers and I was hoping to digest it for my class. The logic made sense to me:
- Instead of clearing denominators every single time (which maybe neither exciting nor informative), why not exploit the that $a_n^2 \geq a_{n-1}a_{n+1}$ for this particular sequence of numbers?
Eventually two other proofs surfaced using other branches of math, but I still don't feel like I got any point across.
In fact, how do I motivate even this one proof? Is this equation really that obvious that it only deserves one sentence?