Prove that $K_1!K_2! \cdots K_n! \geq \left[\frac{K}{n}\right]!^n$

Question

Let $K_1,\ldots,K_n$ be nonnegative integers. Prove that $$K_1!K_2! \cdots K_n! \geq \left[\dfrac{K}{n}\right]!^n,$$ where $K = K_1+\cdots+K_n$.

I was thinking of proving it by induction but there might be an easier way of solving it. How do we deal with the greatest integer less than part?

Answer 1

HINT: Fix the sum $K$, and let's try to find the minimum possible value of $K_1! \cdots K_n!$. So suppose this minimum is attained. We can prove that $|K_i - K_j| \le 1$ for all $i,j$. If this were not the case, then there is some $i,j$ with $K_i > K_j + 1$. Show that subtracting $1$ from $K_i$ and adding $1$ to $K_j$ preserves the sum $K$, but gives a smaller product of factorials. This contradicts the assumption that we had the minimum.

So it's enough to prove the statement in the case $|K_i - K_j| \le 1$ for all $i,j$. If we let $m = \min\{K_1,\ldots,K_n\}$, show that $m = [K/n]$, and that $K_1!\cdots K_n! \ge (m!)^n$.

Answer 2

If you can draw a picture of these whole number it would be so clear the inequality holds.

Let's assume $K$ is divisible by n, then $K/n$ is an integer. And then we can see the numbers in the LHS multiplication are $(K_1+K_1+...+K_n)$ in total, and the numbers in the RHS are $(K/n)*n=K=K_1+K_1+...+K_n$. So the numbers of multipliers on both sides are the same, and if we assume $K_1<=K_2<=...<=K_n$, and you can verify that after we offset numbers on both sides, the remaining numbers on the LHS are greater that that of RHS.

In fact, you can draw a line as $K/n$ in the middle, and the remaining numbers on the LHS are above the line, the remaining ones on RHS are below the line.

And we can do refinement to solve the indivisible case, which might be easy I guess.

Answer 3

There is a real-valued refinement of the inequality, if $x!$ is interpreted to mean $\Gamma(x+1)$, and then it is exactly the statement that $\Gamma(x)$ is log-convex:

$$ K_1 ! K_2 ! \dots K_n! \geq ((\frac{K}{n})!)^n $$

The integer inequality then follows from $\frac{K}{n} \geq {\lfloor \frac{K}{n} \rfloor}$ and $\Gamma(x)$ being an increasing function.