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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 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

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

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

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

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

How about using another method? · 2 years ago

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

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

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. · 1 year, 10 months ago

Nice :). Sorry nowadays I can't come on hangouts. Tell other guys also. · 1 year, 10 months ago

Yup! I will say. :D · 1 year, 10 months ago