Show that $ \binom{n}{i\;j\;k} \le \binom{n}{m\;m\;m} $

Question

Is it true that for integers $i+j+k= 3m = n$ where $i , j, k , m , n\ge 0$ the inequality holds ? $$ \binom{n}{i\;j\;k} \le \binom{n}{m\;m\;m} $$ I tried to show $$ \frac{n!}{m!m!m!} \Big/ \frac{n!}{i!j!k!} = \frac{i!j!k!}{m!m!m!} \ge 1 $$ but I have no idea how to proceed , are there any related theorems ?

Answer 1

It suffices to prove that $i!j!k!\geq (m!)^3$, as you have noted. Consider using Jensens inequality on the convex function $\ln(x!)$: $$f(\frac1n\sum_{k=0}^nx_n)\leq \frac1n\sum_{k=0}^nf(x_n)$$ $$\ln((\frac{i+j+k}3)!)\leq \frac13(\ln(i!)+\ln(j!)+\ln(k!))$$ $$3\ln(m!)\leq \ln(i!j!k!)$$ $$(m!)^3\leq i!j!k!$$ Proven as required. Note that the concavity of $\ln(x!)$ can be quickly proven via the definition of a factorial.

Answer 2

Sketch of the proof...

Start with $(i+1)!(j-1)!=i!j!\frac{i+1}{j}< i!j!$ if $i+1<j$. thus, whenever some of the numbers $i,j,k$ differ by at least $2$, you can replace them with (smaller $+1$, bigger $-1$), without changing the total sum $i+j+k$, and get the product of the factorials smaller. So the product of factorials is the smallest (for the given sum $n=3m$) when the difference between any of $i, j, k$ is at most $1$.

Now note that sum of three numbers $i, j, k$ is divisible by $3$ only if either they all give the same remainder in division by $3$ or all give different remainders. The second case implies that at least two of them would differ by at least two. The first case, together with all them having differences between each other at most $1$, implies that they are all equal. Thus, the smallest product $i!j!k!$ with $i+j+k=n=3m$ occurs when $i=j=k=m$.