There is a typing mistake... I cant make out that stepThe 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
I think you are asking how they get that line from what is given above it.I can't [?] make out the step
" Adding , we have
No no... I understand now.. thanks.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?