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Thread: Prove that

  1. May 20th 2010, 11:46 PM #1
    dhiab
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    Prove that

    If a , b and c are three positf real numbers and :
    <br /> x=\frac{a}{a+b}<br />
    <br /> y=\frac{b}{b+c}<br />
    <br /> z=\frac{c}{c+a}<br />
    Prove that : x+y+z> 1
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  2. May 21st 2010, 12:21 AM #2
    pickslides
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    <br /> <br /> \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=\dots<br /> ??
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  3. May 21st 2010, 12:47 AM #3
    aman_cc
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    Without the loss of generality we can assume
    a >= b >= c > 0

    Under this what can you say about x,y,z ?

    Quote Originally Posted by dhiab View Post
    If a , b and c are three positf real numbers and :
    <br /> x=\frac{a}{a+b}<br />
    <br /> y=\frac{b}{b+c}<br />
    <br /> z=\frac{c}{c+a}<br />
    Prove that : x+y+z> 1
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  4. May 21st 2010, 05:48 AM #4
    roninpro
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    It may also help to scale the numbers so that a+b+c=1 and sharpen the lower bound to x+y+z\geq \frac{3}{2}.
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