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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 $\displaystyle \: \displaystyle (1+e^{1})\cdot (1+e^{2}) \cdot ... \cdot (1+e^{n}) < \displaystyle (e^{1^2 }+e^{2^2 }+e^{3^2}+...+e^{n^2 } )^{\displaystyle 1} \; ,n >1$ . eeuler 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. 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...


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