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Topic: Inequality
Replies: 2   Last Post: Aug 5, 2011 8:38 AM

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Don Coppersmith

Posts: 51
Registered: 2/2/06
Re: Inequality
Posted: Aug 1, 2011 1:33 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

It is easier to prove the stronger statement, where the
multiplier is 1 instead of 2^(1-r):

|a^r - b^r| <= |a - b|^r

Can you prove that one?

Don Coppersmith

> Let a,b >=0 and 0 < r <=1. The inequality (a^r + b^r)
> <= 2^(1-r) .(a + b)^r holds (the function f(x) = x^r
> is concave). How can be proved that:
>
> |a^r - b^r| <= 2^(1-r) .|a - b|^r ?



Date Subject Author
7/31/11
Read Inequality
BobbyG
8/1/11
Read Re: Inequality
Don Coppersmith
8/5/11
Read Re: Inequality
BobbyG

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