If $x+y+z=3k$, where $x, y, z, k$ are integers, prove that $x!y!z! \geq (k!)^3$ 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...
 A: Lemma. If $a,b$ are integers with $0\le a<b$ then $a!b!\le(a+1)!(b-1)!$ with equality iff $b=a+1$.
Proof. Indeed, $a!b! = a!(b-1)!\cdot b\stackrel{(*)}\ge a!(b-1)!\cdot(a+1)=(a+1)!(b-1)!$ and equality at $(*)$ holds iff $b=a+1$. $\square$
Fix $k\in\Bbb N$ and consider 
$$ A:=\{\,x!y!z!\mid x,y,z\in \Bbb N_0, x+y+z=3k\,\}.$$
As a finite set, $A$ certainly has a maximal element. Pick $x,y,z\in \Bbb N_0$ with $x+y+z=3k$ and $x!y!z!=\max A$.
Wlog. $x\le y\le z$. Assume $x<z$.
Then by the lemma $$(x+1)!y!(z-1)! \ge x!y!z!=\max A$$
and hence $(x+1)!y!(z-1)! = x!y!z!$, which - again according to the lemma - means $z=x+1$. If $x\ge k$, we find $3k=x+y+z\ge k+k+(k+1)$, contradiction; and if $x<k$, we find $3k=x+y+z\le (k-1)+k+k$, again a contradiction. We conclude that $x=z$ and then $x=y=z=k$.
A: 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}
$. 
A: $x = k + \alpha,y = k + \beta, z = k+\gamma$
$\alpha + \beta + \gamma = 0$
Without loss of generality we can insist that $\alpha \leq \beta \leq \gamma $
Suppose, $\beta < 0$.
$\alpha +\beta = - \gamma$
$x!y!z! = (k!)^3 \dfrac{(k+1)(k+2)...(k+\alpha)(k+\alpha+1)...(k+\gamma)}{(k-1)(k-2)...(k-\alpha)(k-1)...(k-\beta)}\geq (k!)^3$
In that fraction there are as many factors in the numerator as in the denomoninator.  Each factor in the numerator is > k, each in the denominator is < k.  We can pair them off anyway we want and the ratio will be greater than 1.
and, if $\beta > 0$ 
$x!y!z! = (k!)^3 \dfrac{(k+1)(k+2)...(k+\beta)(k+1)...(k+\gamma)}{(k-1)(k-2)...(k-\alpha)(k-\alpha-1)...(k-\alpha)}\geq (k!)^3$
and if $\beta = 0$
$x!y!z! = (k!)^3 \dfrac{(k+1)(k+2)...(k+\gamma)}{(k-1)(k-2)...(k-\alpha)}\geq (k!)^3$
