Can You Prove This Factorial Inequality for All Positive Integers?

[anemone]

Here is this week's POTW:

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Prove that $$\frac{x!}{x^x}\le \frac{1}{2^{x-1}}$$ for all positive integers $x$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[MarkFL]

Hello MHB Community! (Wave)

I am going to stand in for anemone for a few weeks.

Congratulations to the following for their correct submissions:

• castor28
• kaliprasad

castor28's solution is as follows:

Spoiler

As the proposition is true for $x=1$, the result will follow by induction if we can prove that $\displaystyle\frac{f(x+1)}{f(x)}\le\frac12$, where $f(x)=\dfrac{x!}{x^x}$.

We have:
$$\begin{align*}
\frac{f(x+1)}{f(x)} &= \frac{(x+1)!\,x^x}{x!\,(x+1)^{x+1}}\\
&= \frac{(x+1)x^x}{(x+1)^{x+1}}\\
&= \left(\frac{x}{x+1}\right)^x\\
&= \frac{1}{\left(1+\frac1x\right)^x}
\end{align*}$$
By the binomial theorem, we have:
$$\left(1+\frac1x\right)^x = 1 + \frac{x}{x} + S$$
where $S=0$ for $x=1$ and $S$ is a sum of positive terms for $x>1$. In any case, we have:
$$\left(1+\frac1x\right)^x = \frac{f(n)}{f(n+1)} \ge2$$
and this completes the proof.

The solution provided to me by anemone is:

Spoiler

For $x=1$, we have $1=1$, which holds.

For $x>2,$ we have

$$\begin{align*}\frac{\frac{1}{x}+\frac{2}{x}+\cdots+\frac{x-1}{x}}{x-1}&\ge \sqrt[x-1]{\frac{(x-1)!}{x^{x-1}}}\text{ (By the AM-GM inequality)}\\\left(\frac{\frac{x(x-1)}{2x}}{x-1}\right)^{x-1}&\ge \frac{(x-1)!}{x^{x-1}}\\\frac{1}{2^{x-1}}&\ge \frac{x!}{x^x} \text{ (Q.E.D.)}\end{align*}$$

Hence, $$\frac{x!}{x^x}\le \frac{1}{2^{x-1}}$$ for all positive integers $x$.