My Math Forum Inequalities #3

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 October 9th, 2019, 08:07 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Inequalities #3 Prove that $\displaystyle \: \displaystyle (1+e^{-1})\cdot (1+e^{-2}) \cdot ... \cdot (1+e^{-n}) < \displaystyle (e^{-1^2 }+e^{-2^2 }+e^{-3^2}+...+e^{-n^2 } )^{\displaystyle -1} \; ,n >1$ . e-euler constant . Method required ! Last edited by idontknow; October 9th, 2019 at 08:14 AM.
 October 9th, 2019, 09:07 AM #2 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle Using Bernoulli inequality: $L>1+e^{-1} + e^{-2}+...+e^{-n}$ Show that, $e^{-n}>e^{-n^2}$. Therefore, $L>1+e^{-1} + e^{-2}+...+e^{-n}>e^{-1^2}+e^{-2^2}+...+e^{-n^2}$ Invert the previous inequality, you get, $L<(e^{-1^2}+e^{-2^2}+...+e^{-n^2})^{-1}$ L stands for the left side of inequality. Thanks from idontknow Last edited by tahirimanov19; October 9th, 2019 at 09:12 AM.
 October 9th, 2019, 09:10 AM #3 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle $(1+x_1)(1+x_2)+...+(1+x_n) \ge 1+x_1+x_2+...+x_n$ $x_1,x_2,...,x_n>-1$ and $sgn(x_1)=sgn(x_2)=...=sgn(x_n)$ ------ Also, for any x>-1, $(1+x)^n \ge 1+nx, \; n \in \mathbb{N}, n>1$. And equality is true iff x=0. ------ Prove these.
 October 9th, 2019, 09:24 AM #4 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle If you want inequalities, I can post plenty of them... Thanks from idontknow

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