Inequalities (1 Viewer)

PatHummels · Aug 9, 2014

$If n!<(n+1)^n$

$Show that \sqrt[n]{n!}<\sqrt[n+1]{(n+1)!}$

sorry for lack of spaces

integral95 · Aug 9, 2014

PatHummels said:
$If n!<(n+1)^n$

$Show that \sqrt[n]{n!}<\sqrt[n+1]{(n+1)!}$

sorry for lack of spaces

$From the first equation, take nth root of both sides $\\ \\ n!^{\frac{1}{n}} < n+1 \\ \\ n!*n!^{\frac{1}{n}} < n!*(n+1)= (n+1)! \\ \\ (n!)^{\frac{n+1}{n}}< (n+1)! \\ \\ $then take the (n+1)th root from both sides$

PatHummels · Aug 9, 2014

Thanks, what about a^4+b^4+c^4+d^4>= 4abcd ?

integral95 · Aug 9, 2014

I don't know if there is a flaw in this

$(a^2-b^2)\geq 0 \Rightarrow a^4+b^4 \geq 2a^2b^2 \\ \\ $Similarly$ \ \ c^4+d^4 \geq2c^2d^2 \\ \\ \therefore a^4+b^4+c^4+d^4\geq 2a^2b^2+2c^2d^2 \\ \\ $now consider$ \ \ (x+y)^2\geq 0 \Rightarrow x^2 +y^2 \geq 2xy \\ $Let x = ab and y = cd \\ \\ \Rightarrow a^2b^2 + c^2d^2\geq 2abcd\\ \\ \therefore a^4+b^4+c^4+d^4\geq 2a^2b^2+2c^2d^2\geq 4abcd$

There's probably better way than that lol

go_cadel · Aug 9, 2014

PatHummels said:
Thanks, what about a^4+b^4+c^4+d^4>= 4abcd ?

The result directly follows from am-gm (if allowed to use it)

aDimitri · Aug 9, 2014

go_cadel said:
The result directly follows from am-gm (if allowed to use it)

never allowed to use it lol

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Inequalities (1 Viewer)

PatHummels

New Member

integral95

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PatHummels

New Member

integral95

Well-Known Member

go_cadel

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aDimitri

i'm the cook

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