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 $n$ .

#Algebra #Sharky #ReverseRearrangement

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`2 \times 3`	$2 \times 3$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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 - 5 years, 10 months ago

Same method. @Sharky Kesa use of AM>=GM is a very common inequality in questions involving "!"

Aditya Kumar - 5 years, 10 months ago

How about using another method?

Sharky Kesa - 5 years, 10 months ago

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 - 5 years, 9 months ago

Nice :). Sorry nowadays I can't come on hangouts. Tell other guys also.

Aditya Kumar - 5 years, 9 months ago

Yup! I will say. :D

Surya Prakash - 5 years, 9 months ago

Yeah I thought about Daniel's Reverse Rearrangement inequality.

A Former Brilliant Member - 5 years, 10 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}`	$\boxed{123}$