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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, 6 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.

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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 \(n=1\).

Mine was similar too.

Mine was similar process.

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TopNewestUsing 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.

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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 \(n=1\).

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Mine was similar too.

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Mine was similar process.

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

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