Inequality question from Real Analysis

  • Thread starter kbrono
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  • #1
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Homework Statement


let n[tex]\in[/tex]N To prove the following inequality

na[tex]^{n-1}[/tex](b-a) < b[tex]^{n}[/tex] - a[tex]^{n}[/tex] < nb[tex]^{n-1}[/tex](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
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi kbrono! :smile:

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

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

what's worrying you about it? :confused:

(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 :wink:)
 
  • #3
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Ok thank you for looking it over!
 
  • #4
The inequality is not complete. For n=1 you get
b-a < b-a < b -a

Which is not true.
 

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