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

1. Oct 24, 2012

Polychoron

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:
$$x^{n}=[y+(x-y)]^{n}=\sum_{k=0}^{n}\binom{n}{k}y^{k}(x-y)^{n-k}=y^{n}+\sum_{k=0}^{n-1}\binom{n}{k}y^{k}(x-y)^{n-k}$$
from here:
$$x^{n}-y^{n}=\sum_{k=0}^{n-1}\binom{n}{k}y^{k}(x-y)^{n-k}=(x-y)[\sum_{k=0}^{n-2}\binom{n}{k}y^{k}(x-y)^{n-k-1}+\binom{n}{n-1}y^{n-1}]>(x-y)ny^{n-1}$$
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:
$$x^{n}-y^{n}=(x-y)[\sum_{k=0}^{n-2}\binom{n}{k}x^{k}(y-x)^{n-k-1}+\binom{n}{n-1}x^{n-1}]$$

Ans it stuck.

2. Oct 24, 2012

mathman

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

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

3. Oct 25, 2012

Polychoron

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: Oct 25, 2012
4. Oct 25, 2012

pwsnafu

It's just a telescoping sum. Did you try?

5. Oct 25, 2012

Polychoron

Ya,
I thought maybe there is a way to derived it directly from xn-yn...

6. Oct 25, 2012

MarneMath

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.

7. Oct 25, 2012

Polychoron

OK, thanks pals :)

8. Oct 25, 2012

mathman

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