# Tchebychev's inequality

#### chandra21

##### Junior Member
O can make out the step
(n-1).....

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#### mmm4444bot

##### Super Moderator
Staff member
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
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
 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 $$\;$$ Click to expand... Ok next time i will do that

#### Dr.Peterson

##### Elite Member
I can't [?] make out the step
(n-1).....
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
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.