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Don't use induction

\[\large \left ( \dfrac {2}{3} \right ) ^ n n! \leq \left ( \dfrac {n + 1}{3} \right )^n\]

Prove the above inequation is true for all positive integers \(n\).

Note by Sharky Kesa
2 years, 2 months ago

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Apply AM -GM on \( 1,2,3...,n \).

AM = \( \dfrac{1 + 2 + 3 + ... + n}{n} = \dfrac{\frac{n(n+1)}{2}}{n} = \dfrac{(n+1)}{2} \)

GM = \( \sqrt[n]{1*2*3*...*n} = \sqrt[n]{n!} \)

Now, AM \( \geq \) GM

\( \implies \dfrac{(n+1)}{2} \geq \sqrt[n]{n!} \)

\( \implies \left (\dfrac{(n+1)}{2} \right )^n \geq n! \)

\( \implies \left ( \dfrac{(n+1)}{3} \right )^n \geq (n!) \left (\dfrac{2}{3} \right )^n \) Siddhartha Srivastava · 2 years, 2 months ago

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@Siddhartha Srivastava Same method. @Sharky Kesa use of AM>=GM is a very common inequality in questions involving "!" Aditya Kumar · 2 years, 2 months ago

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@Aditya Kumar How about using another method? Sharky Kesa · 2 years, 2 months ago

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By Reverse Rearrangement inequality,

\[\Large \left( \dfrac{1}{3} + \dfrac{1}{3} \right) \left( \dfrac{2}{3} + \dfrac{2}{3} \right) \ldots \left( \dfrac{n}{3} + \dfrac{n}{3} \right) \]

\[\Large \leq \left( \dfrac{1}{3} + \dfrac{n}{3} \right) \left( \dfrac{2}{3} + \dfrac{n-1}{3} \right) \ldots \left( \dfrac{n}{3} + \dfrac{1}{3} \right)\]

So, on simplification the above inequality reduces to

\[\Large \left( \dfrac{2}{3} \right) ^{n} n! \leq \left( \dfrac{n+1}{3} \right) ^ {n} \]

which is our desired result. Surya Prakash · 2 years ago

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@Surya Prakash Nice :). Sorry nowadays I can't come on hangouts. Tell other guys also. Aditya Kumar · 2 years ago

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@Aditya Kumar Yup! I will say. :D Surya Prakash · 2 years ago

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Yeah I thought about Daniel's Reverse Rearrangement inequality. Chinmay Sangawadekar · 2 years, 1 month ago

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