Binomial coefficients inequality It seems to me that there should be a simple way to prove that
$$
\binom{n}{s+1+a} + \binom{n}{a} \leq \binom{n}{s}
$$
For $s > n/2$ and $a < n-s$.
However it looks like I'm missing it. Any suggestions?
 A: I proved it a couple of month ago, but as no one else provided an answer I'll just write my own. I'll prove the following claim:
$$
\binom{n}{k-1-a} + \binom{n}{a} \leq \binom{n}{k}
$$
for every $ k \leq n/2$ and every $0\leq a < k$ (this is the same as the original proposition but with $k = n-s$).
proof
Because $\binom{n}{k-1-a} + \binom{n}{a}$ is symmetric for $a$ around $(k-1)/2$ we suppose $a \leq (k-1)/2$. We have
$$
{n \choose k}-{n \choose k-1}={n \choose a}\frac{(n-a)\cdots(n-k+2)(n-2k+1)}{k\cdots(a+1)}
$$
and since $\frac{n-a}{k}\geq1.5$ and $\frac{n-2k+1}{a+1}\geq\frac{2}{k+1}$
we get that
$$
{n \choose k}-{n \choose k-1}\geq{n \choose a}(1.5)^{k-a-1}\frac{2}{k+1}\geq{n \choose a}(1.5)^{(k-1)/2}\frac{2}{k+1}
$$
and since $(1.5)^{(k-1)/2}\frac{2}{k+1}\geq1$ for $k\geq9$ we get
that
$$
{n \choose k}\geq{n \choose k-1}+{n \choose a}\geq{n \choose k-1-a}+{n \choose a}\;.
$$
If $k\leq8$ and $n\geq21$ we get that $\frac{n-2k+1}{a+1}\geq1$
and the stronger claim, ${n \choose k}-{n \choose k-1} \geq {n \choose a}$,
holds. For $n<21$ we can easily verify using a computer that the
proposition (not the stronger claim) indeed holds.
