Maximum of $\frac{n!}{i!j!k!}$ under $i+j+k=n$ For a positive integer $n$ and nonnegative inergers $i,j,k \in \mathbb{Z}_{\ge 0}$ with $i+j+k=n$, we define
\begin{align*}
a_{i,j,k}=\frac{n!}{i!j!k!}.
\end{align*}
Can we obtain the maximum of $a_{i,j,k}$?
If $n=3m$ ($m$ is a positive integer), we have
\begin{align*}
\max_{\substack{i,j,k \in \mathbb{Z}_{\ge 0},\\i+j+k=3m}}a_{i,j,k}=a_{m,m,m}.
\end{align*}
Indeed, we have
\begin{align*}
\max_{\substack{i,j,k \in \mathbb{Z}_{\ge 0},\\i+j+k=3m}}a_{i,j,k} =\max_{\substack{i,j,k \in \mathbb{Z}_{\ge 0},\\i+j+k=3m,\,i\le j \le k}}a_{i,j,k}. 
\end{align*}
Under the conditions, $i,j,k \in \mathbb{Z}_{\ge 0}$, $i+j+k=3m$, and $i\le j \le k$, the minimum value of $k$ is $m$. This shows that the minimum value of $j$ is $m$. Accordingly, we have $i!j!k! \ge (m!)^3$. Therefore, $\max_{\substack{i,j,k \in \mathbb{Z}_{\ge 0},\\i+j+k=3m}}a_{i,j,k}=a_{m,m,m}$.
When $n=3m+1$, can we prove that $\max_{\substack{i,j,k \in \mathbb{Z}_{\ge 0},\\i+j+k=3m}}a_{i,j,k}=a_{m,m,m+1}$? If you know a simple proof, please let me know.
Thank you in advance.
 A: Without loss of generality suppose $i\ge j+2$. If $(i,j)$ are replaced with $(i-1,j+1)$, the denominator is multiplied by $\frac{j+1}i\le\frac{j+1}{j+2}<1$ and the whole expression increases.
From this it is easy to see that the whole expression is maximised when $i,j,k$ are as close as possible (differ by at most $1$) à la Turán graphs, a result generalising to any $n$ and any number of denominator factorials. You are right in saying that for $n=3m+1$ the expression is maximised for $(m,m,m+1)$.
A: It's not hard to see that $a_{i, j, k} = \binom{n}{i}\binom{n-i}{j}$. Notice that for binomial coefficient $\binom{a}{b}$, we get the maximum value when $b = \lfloor \frac{a}{2} \rfloor \approx \frac{a}{2}$. So let's fix $i$ in $a_{i, j, k}$, then the maximum value of $\binom{n-i}{j}$ occurs when $j \approx \frac{n-i}{2}$. Then because $i + j + k = n$, we also have $k \approx \frac{n-i}{2}$. Now change your viewpoint and write $a_{i, j, k} = \binom{n}{j}\binom{n - j}{i}$. Then the maximum value of the binomial coefficient $\binom{n - j}{i}$ occurs for $i \approx \frac{n - j}{2}$ and $k \approx \frac{n - j}{2}$.
So If we select each number $i$, $j$, or $k$, then the other two numbers must be equal (or as close as possible). So for $n = 3m$ we must have $(i, j, k) = (m, m, m)$, for $n = 3m + 1$ we must have $(i, j, k) = (m, m, m + 1)$ and if $n = 3m + 2$, then $(i, j, k) = (m, m + 1, m + 1)$.
