Inequality question from Real Analysis

[kbrono]

Homework Statement

let n$$\in$$N To prove the following inequality

na$$^{n-1}$$(b-a) < b$$^{n}$$ - a$$^{n}$$ < nb$$^{n-1}$$(b-a)

0<a<b

Homework Equations

The Attempt at a Solution

Knowing that b^n - a^n = (b-a)(b^(n-1) + ab^(n-2) + ... + ba^(n-2) + a^(n-1) we can divide out (b-a) because b-a # 0. so we have

n(a^(n-1)) < (b^(n-1) + ab^(n-2) + ... + ba^(n-2) + a^(n-1) < n(a^(n-1))

and in the middle of the inequality we see that

na^(n-1) < a^(n-1)< ab^(n-2) + ... + ba^(n-2) < a^(n-1) <nb^(n-1)

and in the middle there are a^j (b^(n-1-j)) terms as j --> n-2
So there are (n-1)+1 terms in the middle or n terms. So if a<b this inequality holds...


I think i pulled some random stuff out of thin air, but its a try.. THanks

 

[tiny-tim]

hi kbrono!

(have a sigma: ∑ and try using the X² icon just above the Reply box )

yes, that's exactly the way you're supposed to do it!

what's worrying you about it?

(btw, you could shorten it a bit by writing ∑_i≤n-1aⁱb^n-1-i …

then ∑_i≤na^n-1 < ∑_i≤n-1aⁱb^n-1-i < ∑_i≤n-1b^n-1 )

 

[kbrono]

Ok thank you for looking it over!

 

[╔(σ_σ)╝]

The inequality is not complete. For n=1 you get
b-a < b-a < b -a

Which is not true.