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Homework Help: Proof this inequality using Chebyshev's sum inequality

  1. Mar 23, 2007 #1
    1. The problem statement, all variables and given/known data

    Let a,b,c,d,e be positive real numbers. Show that

    [tex]\displaystyle{\frac{a}{b+c}} + \displaystyle{\frac{b}{c+d}} + \displaystyle{\frac{c}{d+e}} + \displaystyle{\frac{d}{e+a}} +\displaystyle{\frac{e}{a+b}} \geq \displaystyle{\frac{5}{2}} [/tex]

    2. Relevant equations

    Chebyshev's sum inequality:

    3. The attempt at a solution

    Assume: a > b > c > d > e

    Then: a+b > a+e > b+c > c+d > d+e

    Or: [tex] \displaystyle{\frac{1}{d+e}} \geq \displaystyle{\frac{1}{c+d}} \geq \displaystyle{\frac{1}{b+c}} \geq \displaystyle{\frac{1}{a+e}} \geq \displaystyle{\frac{1}{a+b}} [/tex]

    Hence, we have:

    [itex] a \geq b \geq c \geq d \geq e \geq[/itex]

    [tex] \displaystyle{\frac{1}{d+e}} \geq \displaystyle{\frac{1}{c+d}} \geq \displaystyle{\frac{1}{b+c}} \geq \displaystyle{\frac{1}{a+e}} \geq \displaystyle{\frac{1}{a+b}} [/tex]

    But...this [tex]\sum a_{k}*b_{k}[/tex] is NOT matching up like it what the question is asking...

    Am I arranging these numbers wrong?

    What I am trying to say is:

    [tex] \displaystyle{\frac{a}{d+e}} + \displaystyle{\frac{b}{c+d}} + \displaystyle{\frac{c}{b+c}} + \displaystyle{\frac{d}{a+e}} + \displaystyle{\frac{e}{a+b}} [/tex]


    [tex]\displaystyle{\frac{a}{b+c}} + \displaystyle{\frac{b}{c+d}} + \displaystyle{\frac{c}{d+e}} + \displaystyle{\frac{d}{e+a}} +\displaystyle{\frac{e}{a+b}} \geq \displaystyle{\frac{5}{2}} [/tex]
    Last edited by a moderator: Mar 23, 2007
  2. jcsd
  3. Mar 23, 2007 #2


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    Science Advisor
    Homework Helper

    First, you can't assume a > b > c > d > e, since the sum on the left is not invariant under arbitrary permutations (it is invariant under cyclic permutations (abcde)k though). Also, how are you getting a+e > b+c? That's not a valid inference. And yes, the thing you end up with doesn't match up anyways.
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