How can we show binomial function is convex without calculus? Let $f(z)=\binom{z}{n},$ where $\binom{z}{n}=\frac{z(z-1)\cdots(z-n+1)}{n!}$ and imagine $n\ge 0$. We can show it is convex when $z\ge n$, for example, by calculating $f''(z)$. In fact, it is true for $z\ge n-1$ as dxiv's answer says. But $z\ge n$ is good enough for combinatorial taste. I wonder could we show it is convex by algebraic or combinatorial methods or any other approaches? We may assume $f$ is defined on reals. But proofs for integers are very appreciated too.
 A: Let $P(z) = n! f(z) = z(z-1)...(z-n+1)$ with $n \ge 1$. Since $n!$ is a strictly positive constant factor, $P(z)$ will have the same convexity properties as $f(z)$.
$P$ is a polynomial of degree $n$ and has $n$ distinct real roots $\{0,1,..,n-1\}$. It follows that its derivative $P'$ will have $n-1$ distinct roots in the interval $(0, n-1)$ and, for $n \ge 2$, $P''$ will have $n-2$ distinct real roots in the same interval. It follows that:


*

*For $n \gt 2$ there will be at least one root of $P''$ in $(0,n-1)$, which means that $P''$ will change sign at least once (and in fact exactly $n-2$ times) in the interval. Therefore $P$ can be neither convex nor concave throughout the entire interval $(0,n-1)$.

*$P''$ has no roots outside $(0,n-1)$, so $P$ will be either concave or convex on each side of the interval. Simple sign considerations show that $P$ is:


*

*convex on $[n-1,\infty)$ for all $n \ge 1$

*concave on $(-\infty,0]$ for $n$ odd, and convex on $(-\infty,0]$ for $n$ even.


A: Interpret $f(z) = \binom zn$ as the number of combinations of $n$ objects that can be selected from a set of $z$ distinct objects
where $n, z \in \mathbb Z$ and $z \geq n \geq 0$.
For $n = 0$, we have $f(z) = 1$, which is a convex function.
For $n \geq 1$, consider a set of $z+1$ distinct objects,
$S_{z+1} = \{a_1, \ldots, a_z, a_{z+1}\}$.
Then $\binom {z+1}n$ is the number of combinations of $n$ objects
selected from $S_{z+1}$.
We can partition these combinations into two sets:


*

*The combinations in which $a_{z+1}$ is selected.

*The combinations in which $a_{z+1}$ is not selected.


The number of combinations in the first partition is the number of ways to select $a_{z+1}$ from $\{a_{z+1}\}$ (namely $1$) times the number of ways to select the remaining $n-1$ elements from the $z$-element set $S_{z+1} \setminus \{a_{z+1}\}$, which is $\binom z{n-1}$. 
So there are $\binom z{n-1}$ combinations in this partition.
The number of combinations in the second partition is the number of ways to select $n$ objects from the $z$-element set $S_{z+1} \setminus \{a_{z+1}\}$, so there are $\binom zn$ combinations in this partition.
Adding together the size of the partitions gives the total number of combinations, so we have $$\binom z{n-1} + \binom zn = \binom {z+1}n.$$
(This is a well-known result but I wanted to make sure it was presented
with a combinatorial explanation.)
It follows that
$$
\binom {z+1}n - \binom zn = \binom z{n-1}
$$
and since $z+1 \geq n$ as well,
$$
\binom {z+2}n - \binom {z+1}n = \binom {z+1}{n-1}.
$$

Since $n - 1 \geq 0$, a similar argument to the one above (including the case $n-1=0$ and the case $n - 1 \geq 1$) shows that 
$$
\binom z{n-1} \leq \binom {z+1}{n-1},
$$
that is,
$$
\binom {z+1}n - \binom zn \leq \binom {z+2}n - \binom {z+1}n.
$$
We can extend this via induction to show that if $n \leq z_i < z_m < z_f$
then
$$
(z_f - z_m) \left( \binom{z_m}n - \binom{z_i}n \right)
 \leq (z_m - z_i) \left( \binom{z_f}n - \binom{z_m}n \right).
$$
It follows that
$$
f(z_m) = \binom{z_m}n
 \leq \frac{z_f - z_m}{z_f - z_i} \binom{z_i}n 
       + \frac{z_m - z_i}{z_f - z_i} \binom{z_f}n
 = (1-\lambda) f(z_i) + \lambda f(z_f)
$$
where $0 < \lambda = \dfrac{z_m - z_i}{z_f - z_i} < 1$,
which satisfies a reasonable definition of what it means for a
function over integers to be convex over the integers greater than or
equal to $n$.
