# 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

## 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

Homework Helper
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!

(btw, you could shorten it a bit by writing ∑i≤n-1aibn-1-i

then ∑i≤nan-1 < ∑i≤n-1aibn-1-i < ∑i≤n-1bn-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.