An Inequality

If $n$ is a positive integer, prove that $n^n(\frac{n+1}{2})^{2n} \geq (n!)^3$

#Algebra #Factorial #Inequality

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Using AM-GM Inequality,
$\displaystyle \frac{ 1^3 + 2^3 + 3^3 + 4^3 + \dots + (n-1)^3 + n^3 }{n} \geq \bigg(1^3 \times 2^3 \times \dots \times n^3 \bigg)^{\frac{1}{n}}$
$\displaystyle \Rightarrow \bigg( \frac{n^2}{n} \big( \frac{n+1}{2} \big)^2 \bigg)^n \geq (1.2.3. \dots .(n-1).n)^3$

$\displaystyle \Rightarrow n^n \big( \frac{n+1}{2} \big)^{2n} \geq (n!)^3$

Hence, Proved.

Sudeep Salgia - 6 years, 11 months ago

By intuition: $n^n\geq n!$ with equality iff $n=1$ . By AM-GM inequality:

$\sum_{k=1}^n k \geq n\sqrt[n]{\prod_{k=1}^n k} ~~\Longrightarrow ~ \dfrac{n(n+1)}{2}\geq n\sqrt[n]{n!}~~\Longrightarrow ~ \left(\dfrac{n+1}{2}\right)^{2n}\geq (n!)^2.$

Multiplying above two inequalities gives:

$n^n \left(\dfrac{n+1}{2}\right)^{2n}\geq n!\times (n!)^2 =(n!)^3$

with equality iff $n=1$ .

Jubayer Nirjhor - 6 years, 11 months ago

Mine was similar process.

Fahim Shahriar Shakkhor - 6 years, 11 months ago

Does the latex symbols look odd (some are small some are big) in your device?

Jubayer Nirjhor - 6 years, 11 months ago

@Jubayer Nirjhor – No..I can see fine.

Fahim Shahriar Shakkhor - 6 years, 11 months ago

Mine was similar too.

Mursalin Habib - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$