Inequalities #3

Dec 2015
1,084
169
Earth
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:
Mar 2015
182
68
Universe 2.71828i3.14159
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:
  • Like
Reactions: 1 person
Mar 2015
182
68
Universe 2.71828i3.14159
$(1+x_1)(1+x_2)+...+(1+x_n) \ge 1+x_1+x_2+...+x_n$

$x_1,x_2,...,x_n>-1$ 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.
 
Mar 2015
182
68
Universe 2.71828i3.14159
If you want inequalities, I can post plenty of them...
 
  • Like
Reactions: 1 person