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Math Help - Fraction Chain Inequality Proof

  1. #1
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    Fraction Chain Inequality Proof

    Prove that

    \frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}  {e+f}+\frac{e}{f+a}+\frac{f}{a+b}\geq 3

    Condition: a, b, c, d, e, f are positive real numbers.
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  2. #2
    Pim
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    Edit: I'm not sure how awake I was when I typed this. This is to remain here purely to remind me of my own stupidity.

    Take the first part:

     <br />
\frac{a}{b+c}+\frac{b}{c+d}<br />

    Add the two by multiplying both with the denominator of the other.

    (b+c)*(c+d) = bc + cd + c^2 + bd

     <br />
\frac{ac + ad}{bc + cd + c^2 + bd} + \frac{bc + b^2}{bc + cd + c^2 + bd} = \frac{ac + ad + bc + b^2}{bc + cd + c^2 + bd} <br />
    bc occurs both in the numerator and denominator, so this simplifies to

     <br />
\frac{ac + ad + bc + b^2}{bc + cd + c^2 + bd} = 1 + \frac{ac + ad + b^2}{ cd + c^2 + bd}<br />

    You can do this with the third and forth as well as the fifth and sixth fractions yielding an expression in the form of: 3 + \frac{...}{...} and because all the numbers are positive and real, this fraction will never produce a negative number.
    Last edited by Pim; December 25th 2008 at 12:27 AM.
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  3. #3
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Pim View Post
    bc occurs both in the numerator and denominator, so this simplifies to

     <br />
\frac{ac + ad + bc + b^2}{bc + cd + c^2 + bd} = 1 + \frac{ac + ad + b^2}{ cd + c^2 + bd}<br />
    Who told you?


    Anyway, the inequality is the Shapiro inequality for n=6. You might find a proof along the same lines as those given for the simpler Nesbitt’s inequality on the Wikipedia page.
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  4. #4
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    Quote Originally Posted by JaneBennet View Post
    Who told you?


    Anyway, the inequality is similar to Nesbitt’s inequality. You might find a proof along the same lines as those given on the Wikipedia page.
    You can't cancel the bc terms like that, right?

    And how does the Nesbitt Inequality extend to this 6-term inequality?
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