Minimum value of factorial multiplication

Question

I studied that if x + y + z = 3n (i.e if value of x+y+z is some fixed number which is multiple of 3) then minimum value of x!y!z! is when x=y=z. How can we prove this statement?

Attempt: Well, I couldn't think of much. I started from here: let x = a, y = b, z = n - (a+b) Then x!y!z! = (a!)(b!)(n-(a+b))! And now I'm stuck.

Answer 1

Proof idea:

Your attempt with $x=a,y=b,z=n-(a+b)$ doesn't work because there's two unknowns, $a,b$, and it's harder to study when a function is minimised over two variables.

Try this with $x+y=2n$ ($x=a,y=2n-a$), and show that the minimum is when $x=y=a=n$.

(The calculation is $x!y!=a!(2n-a)!$, and then consider what happens when $a$ increases by $1$: we get an additional $(a+1)$ term in the first factorial and remove a $(2n-a)$ term in the second, so the value will increase by a factor of $\frac{a+1}{2n-a}$. When is this greater than, equal to, and less than $1$?)

Then, this tells you that if $z=c$ and $c$ is even, then $x!y!z!$ is minimum when $x=y=(3n-c)/2$. Given this, show that the value of $c$ to minimise the overall function is $c=n$.

After this, there's a little edge case work to do in showing what happens if $c$ is odd and why this cannot give a smaller answer.

Answer 2

I don't think any proof here is required. What you actually need are two things:

1. $\Gamma(x)$ is logarithmically convex. [Reference?]
2. Jensen's inequality (here you need its particular case: $\log \Gamma((a+b+c))/3)\leq (\log\Gamma(a) + \log \Gamma(b) + \log \Gamma(c))/3$, where $a = x + 1$, etc)