Looking for an idea for proving inequality, probably using binomial theorem.

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Discussion Overview

The discussion revolves around proving a specific inequality involving real numbers and natural numbers, specifically the inequality that states if x > y > 0 and n > 1, then n(x - y)yn-1 < xn - yn < n(x - y)xn-1. The conversation includes attempts to prove both sides of the inequality, with a focus on using the binomial theorem and other mathematical techniques.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a proof for the left part of the inequality using the binomial theorem, showing that x^n - y^n can be expressed as a sum involving (x - y) and y.
  • Another participant suggests that the right part of the inequality can be derived from the identity x^n - y^n = (x - y)(x^(n-1) + x^(n-2)y + ... + y^(n-1), but questions the need for proving this identity.
  • A different participant mentions that the identity could be proven using a telescoping sum, prompting further discussion about the validity of the identity.
  • There is a suggestion to use the polynomial remainder theorem or induction as alternative methods to prove the identity, with some participants expressing that the proof seems straightforward.
  • One participant expresses a preference for deriving identities rather than accepting them without proof, indicating a desire for a more rigorous approach.

Areas of Agreement / Disagreement

Participants generally agree that if the identity is correct, the inequalities follow from it. However, there is no consensus on how to prove the identity itself, with multiple approaches suggested and some participants expressing skepticism about accepting the identity without a proof.

Contextual Notes

Some participants note that the proof of the identity may rely on established mathematical concepts that are not fully explored in the discussion, leading to potential gaps in understanding or assumptions that are not explicitly stated.

Who May Find This Useful

Readers interested in mathematical proofs, particularly those involving inequalities and the binomial theorem, may find this discussion relevant.

Polychoron
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Guess what? I just got my new calculus book last week! ^^

The book opens with the definition of the real numbers by Dedekind and goes to prove properties of this numbering system such as The supremum axiom and others.
At the end of the chapter are about 30 exercises without their solutions; some practice of the chapter and some algebraic preparation for the next chapter deals with limits (mainly inequalities).

I want to consult with you about one inequality:
Prove that if x>y>0 and n>1 natural number than
n(x-y)yn-1< xn - yn < n(x-y)xn-1

I managed to elegantly prove the left part and it goes like that:
[tex]x^{n}=[y+(x-y)]^{n}=\sum_{k=0}^{n}\binom{n}{k}y^{k}(x-y)^{n-k}=y^{n}+\sum_{k=0}^{n-1}\binom{n}{k}y^{k}(x-y)^{n-k}[/tex]
from here:
[tex]x^{n}-y^{n}=\sum_{k=0}^{n-1}\binom{n}{k}y^{k}(x-y)^{n-k}=(x-y)[\sum_{k=0}^{n-2}\binom{n}{k}y^{k}(x-y)^{n-k-1}+\binom{n}{n-1}y^{n-1}]>(x-y)ny^{n-1}[/tex]
As we were asked to show.

About the right inequality, well, it's not working so nicely...
When I try the same concept but slightly different, I have a hard time showing the last transition (where the "<" sine comes to play)

I got:
[tex]x^{n}-y^{n}=(x-y)[\sum_{k=0}^{n-2}\binom{n}{k}x^{k}(y-x)^{n-k-1}+\binom{n}{n-1}x^{n-1}][/tex]

Ans it stuck.
 
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xn - yn = (x-y)(xn-1 + xn-2y +... + xyn-2 + yn-1)

Since there are n terms inside the parenthesis, the inequalities follow immediately.
 
Nice.

I agree that if the identity is right then the inequalities immediately derive from it. But can you prove it?
Of course I can see it by open the parenthesis, I simply don't fan of identities that came out of nowhere...
 
Last edited:
Polychoron said:
But can you prove it?

It's just a telescoping sum. Did you try?
 
Ya,
Polychoron said:
Of course I can see it by open the parenthesis, I simply don't fan of identities that came out of nowhere...
I thought maybe there is a way to derived it directly from xn-yn...
 
I guess if you wanted to prove what mathman used you could use the polynomial reminder theorem, but it's fairly obvious and I think common knowledge. Perhaps induction would be easier, either way, it's straight forward enough, if you desire to do so.
 
OK, thanks pals :)
 
Polychoron said:
Nice.

I agree that if the identity is right then the inequalities immediately derive from it. But can you prove it?
Of course I can see it by open the parenthesis, I simply don't fan of identities that came out of nowhere...

Use the geometric series summation formula with a = xn-1 and r = y/x
 

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