Tchebychev's inequality

[chandra21]

O can make out the step
" Adding , we have
(n-1).....
Please explain

 

[mmm4444bot]

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]

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]

					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]

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 [MATH]a_i b_i + a_j b_j \ge a_i b_j + a_j b_i[/MATH] with [MATH]i\ne j[/MATH], and the inequality stated follows. On the LHS, each product [MATH]a_i b_i[/MATH] 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 [MATH]a_i b_j[/MATH] appears once, so it is just a sum of those.

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

 

[chandra21]

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 [MATH]a_i b_i + a_j b_j \ge a_i b_j + a_j b_i[/MATH] with [MATH]i\ne j[/MATH], and the inequality stated follows. On the LHS, each product [MATH]a_i b_i[/MATH] 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 [MATH]a_i b_j[/MATH] 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.