Maximizing the trinomial coefficient Let $\binom{n} {a,b,c} = \frac{n!}{a!\cdot b! \cdot c!}$
In other words, $\binom{n} {a,b,c}$ is the trinomial coefficient.
I am trying to find the triplets $(a,b,c)$ which maximize this trinomial coefficient.
I have determined, by simply plugging in numbers, that:
When $n\bmod 3 = 0$, the trinomial coefficient is maximized when $a = b = c = \frac{n}{3}$
When $n \bmod 3 = 1$, the trinomial coefficient is maximized when $a = b = \frac{(n - 1)}{3}  \space \text{and} \space c = \frac{(n + 2)}{3} \impliedby$  Note for this case, any two variables can equal $\frac{(n - 1)}{3}$  while the other equals $\frac{(n + 2)}{3}.$
And when $n \bmod 3 = 2$, the trinomial coefficient is maximized when $a = \frac{(n - 2)}{3}$ and $b = c = \frac{(n + 1)}{3} \impliedby$ Same caveat about the interchanging of variables as $n \bmod 3 = 1$.
While I understand this, I cannot think of a way to rigorously prove this. Could someone help me get started in the right direction?

Temporarily fix $c$. We want to minimize $x!y!$ given that $x+y=n-c=k$. 
Suppose that we have $x\lt y$. Compare $x!y!$ with $(x+1)!(y-1)!$.  We have 
$$(x+1)!(y-1)!=\frac{x+1}{y}(x!y!).$$
Thus $(x+1)!(y-1)!\lt x!y!$ unless $y=x+1$, in which case we have equality.
So given $a,b,c$ with fixed sum, if two of $a,b,c$ differ by $2$ or more, we can decrease the product. It follows that the minimum product is reached when at least two of $a,b,c$ are equal, and the third differs from them by at most $1$.
Your observations now follow. Suppose for example that $n$ is divisible by $3$. The product $a!b!c!$ is minimized, and therefore $\frac{n!}{a!b!c!}$ is maximized, when at least two of $a,b,c$ are equal, say $a=b$, and the third is $a$, $a+1$, or $a-1$. The sum is divisible by $3$ precisely if they are all equal.
