# An Inequality

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

Note by Fahim Shahriar Shakkhor
3 years, 11 months ago

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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.

- 3 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 $$n=1$$.

- 3 years, 11 months ago

Mine was similar too.

- 3 years, 11 months ago

Mine was similar process.

- 3 years, 11 months ago

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

- 3 years, 11 months ago

No..I can see fine.

- 3 years, 11 months ago