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Math Help - Inequality Problem

  1. #1
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    Inequality Problem

    happy bithday for all.
    I hope you that help me to prove this inequalitie:
    a,b,c are positiv reel
    sqrt(2a/(a+b) + sqrt(2b/(b+c)+ sqrt(2c/(c+a)<3.
    Thank you in advance.
    Last edited by janvdl; January 5th 2011 at 09:48 AM. Reason: Fixed title
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  2. #2
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    Suppose that each of a,~b,~\&~c is a positive real numbers.
    If a=b=c then \sqrt{\dfrac{2a}{a+b}}+\sqrt{\dfrac{2b}{b+c}} +\sqrt{\dfrac{2c}{a+c}}=3
    If we have a<b= c then \sqrt{\dfrac{2a}{a+b}}+\sqrt{\dfrac{2b}{b+c}} +\sqrt{\dfrac{2c}{a+c}}<3 .
    WHY?
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  3. #3
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    Thank you Platon.
    I have an idea :
    a/2<(a+b)/2 and 2/(a+b)<2/a and sqrt(2a/(a+b))<sqrt(2).
    by this method we arrive at: sqrt(2a/(a+b))+sqrt(2b/(c+b))+sqrt(2c/(a+c)) <3sqrt(2).
    Thank you again.
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  4. #4
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    Quote Originally Posted by Lamalif View Post
    Thank you Platon.
    I have an idea :
    a/2<(a+b)/2 and 2/(a+b)<2/a and sqrt(2a/(a+b))<sqrt(2).
    by this method we arrive at: sqrt(2a/(a+b))+sqrt(2b/(c+b))+sqrt(2c/(a+c)) <3sqrt(2).
    Thank you again.
    Granted, but I thought you wanted to prove that the expression is less than 3 (not 3\sqrt{2}).
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  5. #5
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    Quote Originally Posted by alexmahone View Post
    Granted, but I thought you wanted to prove that the expression is less than 3 (not 3\sqrt{2}).
    Yes .you are right. But I did not arrive.</span>
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