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

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

## closed as off-topic by Gibbs, Arnaud D., StubbornAtom, Michael Hoppe, GNUSupporter 8964民主女神 地下教會Oct 18 '18 at 11:13

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Gibbs, Arnaud D., StubbornAtom, Michael Hoppe, GNUSupporter 8964民主女神 地下教會
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• $$\dfrac{r+(n+1-r)}2\ge?$$ for $r>0,n+1-r>0$ – lab bhattacharjee Oct 18 '18 at 8:58
• kindly include your attempt if possible. – Siong Thye Goh Oct 18 '18 at 8:58
• 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 9:25
• I've already voted to close as "missing context", but it's also a duplicate of this – Arnaud D. Oct 18 '18 at 9:43

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?

• I would appreciate if you continue – custumuzer Oct 18 '18 at 9:08
• @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 9:09
• @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 9:20
• Beautiful solution! +1. – Michael Rozenberg Oct 18 '18 at 11:09

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.

• I've fixed your equtions with MathJax, but the math is just wrong. – Arnaud D. Oct 18 '18 at 9:32
• Thanks @ArnaudD. But where is the Math wrong? – nature1729 Oct 18 '18 at 9:34
• 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 9:36
• 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 9:39

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

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

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.