Take the 2-minute tour ×  I need to prove such a formula:  $$
\left\{ \begin{matrix} n \\ k-1 \end{matrix} \right\} \cdot
\left\{ \begin{matrix} n \\ k+1 \end{matrix} \right\} \le
\left\{ \begin{matrix} n \\ k \end{matrix} \right\}^{2} \\ $$
Where {} are Stirling numbers of the second kind. (number of
ways to partition a set of n objects into k non-empty subsets).
I tried to figure out some combinatorial proof, but failed.  I'd
be grateful for any help.  share|improve this question What is
the source of this problem? –  Aryabhata Mar 8 '12 at 21:55 A
sequence $c_n$ is called discrete log-concave if $c_n^2
\geqslant c_{n-1} c_{n+1}$. This article by Sibuya proves the
discrete log-concavity for Stirling numbers. –  Sasha Mar 8 '12
at 22:26 Possibly related: An Inequality Involving Bell Numbers
–  Mike Spivey Mar 8 '12 at 22:34 There is a fairly short but
very non-combinatorial proof in Herbert S. Wilf’s
generatingfunctionology, which may be downloaded here as a
hyperlinked PDF. The desired result is Corollary