TITLE: If $x+y+z=3k$, where $x, y, z, k$ are integers, prove that $x!y!z! \geq (k!)^3$
QUESTION [5 upvotes]: If $x+y+z=3k$, where $x, y, z, k$ are integers, prove that 
$x!y!z! \geq (k!)^3$
Well I was able to prove this intuitively, but what i need is a rigorous mathematical proof. I shall explain my intuitive proof.
Let us take $L=x!y!z!$ at at $x=y=z=k$. Thus, $L=(k!)^3$. Now, to attain any other case, you would have to multiply by a number greater than $k$ and divide by a number less that $k$. Which in short means we would have to multiply by a number greater than $1$. This would imply that it's  magnitude would become greater than $L$. Thus using this algorithm, all cases can be obtained and they would all be greater than or equal to $L$. Thus, $L$ is the lowest possible value. 
Well, this is easy to explain. The problem would be to write it down as a proper mathematical proof.
Please help me achieve that...

REPLY [2 votes]: Much more is true.
If $(x_i)$ is a set of $n$ reals
with $x_i \ge 1$,
then
$\left(\frac1{n}\sum x_i\right)!
\le (\prod x_i!)^{1/n}
$,
or
$\left(\frac1{n}\sum x_i\right)!^n
\le \prod x_i!
$.
This is because
the factorial function
is log-convex.
Here is the proof.
Jensen's inequality states
that if $f$ is a convex function
($f''(x) \ge 0$)
then
$f(\frac1{n}\sum x_i)
\le \frac1{n}\sum f(x_i)
$. 
The Gamma function 
and the factorial function
are log-convex -
their logs are convex.
See here for a typical discussion:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.456.1448&rep=rep1&type=pdf
Letting
$f(x)
=\log(x!)
$,
$\log((\frac1{n}\sum x_i)!)
\le \frac1{n}\sum \log(x_i!)
$. 
Exponentiating,
$(\frac1{n}\sum x_i)!
\le (\prod x_i!)^{1/n}
$.