Proof of $ n! \lt \Big(\frac{n+1}{2}\Big)^n$

Question

Use $AM $-$GM$ inequality to show that $$(\forall n \in \mathbb{N}) : n! \lt \Bigg(\frac{n+1}{2}\Bigg)^n$$ .

Your question would be much better if you provided context. I strongly encourage you to read the Q/A I linked to and edit your question accordingly. For example, you don't even tell us where does your question come from, or if you know what AM-GM inequality is, how you've tried to apply it... — Arnaud D., Oct 18 '18 at 09:25
I've already voted to close as "missing context", but it's also a duplicate of this — Arnaud D., Oct 18 '18 at 09:43

score 5 · Answer 1 · answered Oct 18 '18 at 09:00

5

We have $$\sqrt[n]{n!}=\sqrt[n]{1\cdot 2\cdot …\cdot n}\le \frac{1+2+...+n}{n}=\frac{n(n+1)}{2n}$$ Can you finish?

answered Oct 18 '18 at 09:00

I would appreciate if you continue – custumuzer Oct 18 '18 at 09:08
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@custumuzer There is nothing to continue: you only need to read carefully the above answer and write half a line more and then you have what you want! – DonAntonio Oct 18 '18 at 09:09
@custumuzer. Try to carefully compare the answer that has been provided to the original question and you will see what needs to be done – Jaynot Oct 18 '18 at 09:20
Beautiful solution! +1. – Michael Rozenberg Oct 18 '18 at 11:09

nature1729 · Answer 2 · 2018-10-18T09:46:59.843

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Inequality problems are all about finding small patterns. Often you can start from some modest equations. Let $a,b$ are positive numbers.

Such that $a+b=m$

Then, $ab=a(m-a)=am-a^2=\frac{m^2}{4}-(a-m/2)^2$

This means $ab$ is maximum when $a=b=m/2$.

Take, $a+b=n+1$

Hence $$1\cdot2\cdots n=(1\cdot n)(2\cdot (n-1))\cdots \left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)\leq \left(\frac{n+1}{2}\right)^n$$

You can easily work out cases when $n$ is even or odd.

edited Oct 18 '18 at 09:46

answered Oct 18 '18 at 09:25

nature1729

I've fixed your equtions with MathJax, but the math is just wrong. – Arnaud D. Oct 18 '18 at 09:32
Thanks @ArnaudD. But where is the Math wrong? – nature1729 Oct 18 '18 at 09:34
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Second equation, $\left(\frac{a+b}{2}\right)^2-\frac{a^2+b^2}{2}=\frac{-a^2+2ab-b^2}{4}=-\left(\frac{a-b}{2}\right)^2$ and not $ab$. (The square on the first term gives you a $4$ in the denominator instead of a $2$) – Arnaud D. Oct 18 '18 at 09:36
Your last equation was also problematic, but I think I see what you're doing now. Except the second factor in the middle should be $2\cdot (n-1)$. – Arnaud D. Oct 18 '18 at 09:39

KM101 · Answer 3 · 2018-10-18T10:29:43.153

$$\sqrt[n]{x_0\cdot x_1\cdot x_2\cdot...\cdot x_n} \leq \frac {x_0+x_1+x_2+...+x_n}{n}$$

$$n! = n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot...\cdot 1$$

Using the AM-GM inequality, we can conclude:

$$\sqrt[n] {n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot...\cdot 1} \leq \frac{1+...+(n-3)+(n-2)+(n-1)+n}{n}$$

Simplify the right hand side.

$$\sqrt[n] {n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot...\cdot 1} \leq n\biggr(\frac{n+1}{2n}\biggr)$$

Simplify again and raise both sides to power $n$.

$$n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot...\cdot 1 \leq \biggr(\frac{n+1}{2}\biggr)^n$$

$$n! \leq \biggr(\frac{n+1}{2}\biggr)^n$$

Proven.

3 Answers3