If $x+y+z=3k$, where $x, y, z, k$ are integers, prove that $x!y!z! \geq (k!)^3$

Question

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...

Answer 1

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 $x=y=z=k$.

Answer 2

$x = k + \alpha,y = k + \beta, z = k+\gamma$

$\alpha + \beta + \gamma = 0$

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 $\beta = 0$

$x!y!z! = (k!)^3 \dfrac{(k+1)(k+2)...(k+\gamma)}{(k-1)(k-2)...(k-\alpha)}\geq (k!)^3$

Answer 3

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} $.