Waste less time on Facebook — follow Brilliant.
×

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
2 years, 10 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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 · 2 years, 10 months ago

Log in to reply

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 · 2 years, 10 months ago

Log in to reply

@Jubayer Nirjhor Mine was similar too. Mursalin Habib · 2 years, 10 months ago

Log in to reply

@Jubayer Nirjhor Mine was similar process. Fahim Shahriar Shakkhor · 2 years, 10 months ago

Log in to reply

@Fahim Shahriar Shakkhor Does the latex symbols look odd (some are small some are big) in your device? Jubayer Nirjhor · 2 years, 10 months ago

Log in to reply

@Jubayer Nirjhor No..I can see fine. Fahim Shahriar Shakkhor · 2 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...