Lower bound of multinomial coefficients?

Question

Multinomial coefficient $=\dfrac{n!}{a_1!\cdot a_2!\cdots a_k!}$, where $n=a_1+a_2+\cdots+a_k$.

So my thoughts are there should be a minimum when the denominator goes to the largest.

I believe there is a maximum for the denominator from my basic knowledge and experience of inequity, like for $a+b=c$, the $\max(ab)$ is reached when $a=b$, but I don't know how to do with the situation here (factorial).

Can anyone give me some help?

Thanks

Edit: Well actually, I'm using this to help me solve a problem. The answers followed here makes me feel I am trying the problem in a wrong way. Here is the question, could you tell how to think it right?

Let n larger or equal than 2. We want to select as many subsets of $[n]$ as possible, without selecting two subsets so that one of them contains the other. Prove that we can always select at least $(2^n-1)/n$ Subsets.

Thanks again

Answer 1

The lower bound of the coefficient is $1$. This happens when $a_i=n$ for some $1\le i\le k$. Then, we have $a_j=0$ for all $j \ne i$.

It must be noted that here $a_j \ge 0$ for all $1\le j\le k$. Thus, no $a_j$ in the denominator can be greater than $n$.

It can be seen that $1$ must be the minimum this way: the multinomial coefficient is the number of ways of arranging $n$ objects where $a_1$ are of one kind, $a_2$ are of a second kind, and so on. As it is the number of ways, the value must always be a positive integer for $n\ge 0$. So, as it can take the value of $1$, it must be the minimum value.

Answer 2

Note that moving from $a_i!a_j!$ with ($a_j\ge1$) to $(a_i+1)!(a_j-1)!$ increases by a factor of $\frac{a_i+1}{a_j}$, which is $>1$ whenever $a_i+1>a_j$. In particular, the denominator can be enlarged whenever there exist two nonzero arguments. Therefore, the maximal denominator can only be the one corresponding to all but one of the $a_i$ equal to zero. This gives the lower bound $1$ for the multinomial.