Proving $(n!)^3

Question

Show the following with high school math: $$(n!)^3<n^n\times\left(\frac{n+1}{2}\right)^{2n}$$

It's given after an article on the Arithmetic Mean-Geometric Mean inequality, so I suppose it's solved using that.

so I suppose it's solved using that . So, what's your progress so far? — Lee David Chung Lin, Feb 16 '20 at 06:20
I formatted your math expressions using MathJax. Please double-check that the revision matches your intention. — Blue, Feb 16 '20 at 06:23

Jacky Chong · Accepted Answer · 2020-02-17T04:18:29.933

5

Recall the AM-GM inequality is given by \begin{align} \sqrt[n]{a_1a_2\ldots a_n} \leq \frac{a_1+a_2+\cdots+a_n}{n} \end{align} then it follows \begin{align} \sqrt[n]{(n!)^3}=\sqrt[n]{1^3\cdot 2^3\cdots n^3} \leq \frac{1^3+2^3+\cdots+n^3}{n} = \frac{1}{n}\left(\frac{n(n+1)}{2}\right)^2 = \frac{n(n+1)^2}{4}. \end{align} Here I used the sum of cubes formula.

edited Feb 17 '20 at 04:18

answered Feb 16 '20 at 06:45

Jacky Chong

25,739

1

do you mean $1^\color{red}{3}+2^\color{red}{3}+...$? – Feb 16 '20 at 10:17
@user715522 Yes. Thanks for pointing out the typo. – Jacky Chong Feb 17 '20 at 04:18
But it is a strict inequality as per the question. – Derek Luna Feb 17 '20 at 04:29
@DerekLuna AM-GM inequality is equality if and only if $a_1=a_2 =\ldots = a_n$ which is not the case here. – Jacky Chong Feb 17 '20 at 04:31

score 0 · Answer 2 · answered Feb 16 '20 at 06:35

0

I am assuming you mean for n>1.

$n! <n^n$ (obvious).

$\frac{(n+1)^2}{4} \ge n\times1$

$\frac{(n+1)^2}{4} \ge (n-1)\times2$.

....

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Now solve individualy for n=odd and even.

In each case you get $n!<(\frac{n+1}{2})^n$

answered Feb 16 '20 at 06:35

aryan bansal

1,923

You could have given this as a hint in comments – Arjun Feb 16 '20 at 06:45
That's what I am saying, you neither provided a hint nor gave a full detailed solution – Arjun Feb 16 '20 at 06:49

Proving $(n!)^3

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