Tchebychev's inequality

mmm4444bot

Super Moderator
Staff member
Joined
Oct 6, 2005
Messages
10,302
The op looks like an Easter Egg Hunt.

I found the step! For readers, it's located about halfway down the page in the proof (See 2nd attachment: Page 8, HIGHER ALGEBRA).

@chandra21 💡 The forum has a Preview button, for double-checking posts before submitting them. Cheers

\(\;\)
 

chandra21

Junior Member
Joined
May 23, 2013
Messages
60
The op looks like an Easter Egg Hunt.

I found the step! For readers, it's located about halfway down the page in the proof (See 2nd attachment: Page 8, HIGHER ALGEBRA).

@chandra21💡 The forum has a Preview button, for double-checking posts before submitting them. Cheers

\(\;\)
There is a typing mistake... I cant make out that step
 

chandra21

Junior Member
Joined
May 23, 2013
Messages
60

The op looks like an Easter Egg Hunt.

I found the step! For readers, it's located about halfway down the page in the proof (See 2nd attachment: Page 8, HIGHER ALGEBRA).

@chandra21💡 The forum has a Preview button, for double-checking posts before submitting them. Cheers

\(\;\)
Ok next time i will do that
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
3,856
I can't [?] make out the step
" Adding , we have
(n-1).....
Please explain
I think you are asking how they get that line from what is given above it.

As they say, they are adding together all possible inequalities of the form \(\displaystyle a_i b_i + a_j b_j \ge a_i b_j + a_j b_i\) with \(\displaystyle i\ne j\), and the inequality stated follows. On the LHS, each product \(\displaystyle a_i b_i\) appears n-1 times (because any i is paired with each of the other n-1 j's), so they can factor that out. On the RHS, each product \(\displaystyle a_i b_j\) appears once, so it is just a sum of those.

Is there a particular part of that that you don't understand?
 

chandra21

Junior Member
Joined
May 23, 2013
Messages
60
I think you are asking how they get that line from what is given above it.

As they say, they are adding together all possible inequalities of the form \(\displaystyle a_i b_i + a_j b_j \ge a_i b_j + a_j b_i\) with \(\displaystyle i\ne j\), and the inequality stated follows. On the LHS, each product \(\displaystyle a_i b_i\) appears n-1 times (because any i is paired with each of the other n-1 j's), so they can factor that out. On the RHS, each product \(\displaystyle a_i b_j\) appears once, so it is just a sum of those.

Is there a particular part of that that you don't understand?
No no... I understand now.. thanks.
 
Top