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Math Help - Prove that

  1. #1
    Super Member dhiab's Avatar
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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. #2
    Master Of Puppets
    pickslides's Avatar
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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. #3
    Super Member
    Joined
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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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
    485
    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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