TITLE: An Inequality Involving Bell Numbers: $B_n^2 \leq B_{n-1}B_{n+1}$
QUESTION [19 upvotes]: The following inequality came up while trying to resolve a conjecture about a certain class of partitions (the context is not particularly enlightening):
$$
B_n^2 \leq B_{n-1}B_{n+1}
$$
for $n \geq 1$, where $B_n$ denotes the $n$th Bell number (i.e. the number of partitions of an $n$-element set). I ran this inequality through Maple for values of $n$ up to 500 or so and did not find a counterexample.
There is no nice closed form for $B_n$, so I was hoping to prove this inequality combinatorially rather than analytically (particularly since the given inequality is just the simplest version of a more general inequality I hope to establish).
Let $P_n$ be the collection of (not number of) all partitions of an $n$-element set. My approach was to find an injection from $P_n \times P_n$ into $P_{n-1} \times P_{n+1}$. Suppose we are to map $(C_1, D_1)$ to $(C_2, D_2)$ and suppose, for convenience, our ground set is the integers from $1$ to $n$.
A natural seeming choice was to choose $C_2$ to be the partition $C_1$ with the element $n$ removed. Since removing $n$ will map many partitions in $P_n$ to the same partition in $P_{n-1}$, we would need somehow to choose $D_2$ in such a way as to retain information about where $n$ was in $C_1$. We have the new element $n+1$ to work with, so perhaps it can be used to "tag" the partitions in some unique way.
I've stressed a combinatorial approach in this post, but I would greatly appreciate any techniques that might be of use in establishing (or refuting) this inequality.

REPLY [6 votes]: Some more references, which will give you some additional proofs if you're interested in tracking them down. 

Asai, Kubo, and Kuo ["Bell numbers, log-concavity, and log-convexity," Acta Applicandae Mathematicae 63 (2000) 79-87] give an upper bound as well as the lower one:


$$1 \leq \frac{B_{n+1}B_{n-1}}{B_n^2} \leq 1 + \frac{1}{n}.$$


(Added: The Bender and Canfield paper mentioned below gives this bound as well.)

Liu and Wang ["On the log-convexity of combinatorial sequences," Advances in Applied Mathematics 39 (2007) 453-476] state 


"The log-convexity of the Bell numbers was first obtained by Engel ["On the average rank of an element in a filter of the partition lattice," Journal of Combinatorial Theory Series A 65 (1994) 67-78] .  Then Bender and Canfield ["Log-concavity and related properties of the cycle index polynomials," Journal of Combinatorial Theory Series A 74 (1996) 57-70] gave a proof by means of the exponential generating function of the Bell numbers.  Another interesting proof is to use Dobinski formula [as in anon's answer].  We can also obtain the log-convexity of the Bell numbers by Proposition 2.3 and the well-known recurrence $$B_{n+1} = \sum_{k=0}^n \binom{n}{k} B_k."$$

Liu and Wang's proposition 2.3 (due to Davenport and Pólya) says 

If $\{x_n\}$ is log-convex, and $z_n = \sum_{k=0}^n \binom{n}{k} x_k$, then $\{z_n\}$ is log-convex as well.  

While at first this may seem circular when applied to the Bell numbers, it's not. Proposition 2.3 says that if $x_{k-1}x_{k+1} \geq x_k^2$ for $1 \leq k \leq n-1$ then $z_{k-1}z_{k+1} \geq z_k^2$ for $1 \leq k \leq n-1$.  With the Bell number recurrence, then, we have $B_{k-1}B_{k+1} \geq B_k^2$ for $1 \leq k \leq n-1$ implying $B_{k}B_{k+2} \geq B_{k+1}^2$ for $1 \leq k \leq n-1$.  Since we can easily check that $B_0 B_2 \geq B_1^2$, this gives us an inductive proof of the log-convexity of the Bell numbers.

(Added 2) Canfield, in "Engel's inequality for Bell numbers" [Journal of
Combinatorial Theory Series A 72 (1995) 184-187], discusses this inequality as well and gives the same proof as in my other answer.