Question about convex optimization with binomial coefficients

Question

I don't have any experience with optimization other than some very basic problems from elementary calculus, but I want to understand a particular claim from Alon and Spencer's The Probabilistic Method.

We are looking to optimize $v$ and we have $a$ and $b$ such that $b = v-a$. We then want to minimize the expression $$ \frac{{a \choose n} + {b \choose n}}{{v \choose n}}.$$ The text says, "As ${y \choose n}$ is convex, this expression is minimized when $a = b$." I am struggling to see 1) that ${y \choose n}$ is convex and 2) that $b = a$ is optimal based on the fact that ${y \choose n}$ is convex.

For the time being, I am not really concerned with being fast/good at optimization (as that's not at all why I am reading the book), I just want to understand the basics of why this is true so that I can understand what's going on if I come across a similar argument later on in the book.

EDIT: Someone linked to this post, which asks the same question. The following hint was given in response:

Compare $\dfrac{\dbinom{y+1}{n} }{\dbinom{y}{n}}$ with $\dfrac{\dbinom{y}{n} }{\dbinom{y-1}{n}}$ for $y \gt n$

I still am unsure why this is helpful.

Answer 1

Strange to declare that $x\mapsto x(x-1)\ldots(x-n+1)$ is convex with so many zeros for $n>2.$ Is it a serious book?

Answer 2

Let $f(x) = x(x-1)\cdots (x-n+1) = \prod_{k=0}^{n-1}(x-k)$ then,

$$f'(x) = f(x) \sum_{k=0}^{n-1}\frac1{x-k}$$

$$f''(x) = -f(x)\sum_{k=0}^{n-1} \frac{1}{(x-k)^2} + f(x)\left(\sum_{k=0}^{n-1}\frac1{x-k}\right)^2 = f(x) \sum_{k=0}^{n-1}\sum_{\ell=0,\,\ell\neq k}^{n-1}\frac1{(x-k)(x-\ell)}\ge 0$$

this proves convexity. Can you finish the other part of the question?