#Help with proving the above inequality
51 messages · Page 1 of 1 (latest)
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Bruh
Well if I remember correctly, it’s that …
Ok for monotonic increasing sequences $a_n, b_n$ we have that $$\sum_{k=1}a_kb_k \geq \sum_{k=1}a_kb_{\pi(k)}$$ for any permutation $\pi$ on ${1,…,n}$
ℝανι-sαη
You can assume WLOG that b>a
Wait so it can't be done through the sequence and series formulae?
I haven't done permutation and combination yet
See if you can get some Information from the fact that a and b are natural numbers and a ≠b
Yeah but how will you separate them?
I guess you have to separate a b times and b a times
Is it necessary to seperate the terms?
I think not
Yeah if ur doing it through am>gm no?
To get a power b terms u have to separate it b times
Like for example if it was a²b² we would write a/2 + a/2 + b/2 + b/2
Then do the am >gm method
Try to do some algebraic manipulations from AM>GM
I mean we can write it as a/b + a/b ... b times
But I'm not getting anything from that method
Unless u need to shift the powers or smth
Well, you can raise terms to some power which preserves the inequality
How though
Full solutions aren't allowed, sadly, you can only get hints
You have to do the work yourself
- Raise terms to appropriate power
- Do manipulations using the fact that a, b ∈ℕ and a ≠ b
- See how you can obtain the terms
I tried algebraic manipulation but I got no results
Can u give an example of the 1st
Like squaring both sides of some sorts
?
If we want to get the terms we have to split a b times and b a times
Such as, 1/ab?
Try ab also
Wait ur using the am>gm relation right?
Alr
Yea
How one can prove it without AM>GM
@brave lake
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