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Math Help - Prove that

  1. #1
    Super Member dhiab's Avatar
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    Prove that

    a\in R^{+},b\in R^{+}
    if : a+b+c=1
    Prove that :
     <br />
\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq 2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})<br />
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  2. #2
    Super Member PaulRS's Avatar
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    Shouldn't it be: c\in \mathbb{R}^+ too?

    We have:
    \sum\frac{1+a}{1-a}=\sum\left(\frac{1+a}{1-a}-1\right)+3=\sum\left(\frac{2a}{1-a}\right)+3=\sum\left(\frac{2a}{b+c}\right)+3 - all the sums are cyclic-

    Thus: \frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq{2\cdot\left(\frac{a}{b}+\frac{b}{c}+\frac{c  }{a}\right)} iff \frac{3}{2}\leq \sum\left(\frac{a}{b}\right) - \sum\left(\frac{a}{b+c}\right) (1)

    Now: \text{RHS} = \sum\left(\frac{a}{b}-\frac{a}{b+c}\right) = \sum\left( \frac{a}{b}\cdot (1-\tfrac{b}{b+c}) \right)=\sum\left( \frac{a}{b}\cdot \frac{c}{b+c} \right)=\sum\left( \frac{ac}{b^2+bc}\right) (2)

    Let x=\frac{a}{b};y=\frac{b}{c};z=\frac{c}{a}, then (2) is equivalent to: RHS=\sum\left(\frac{x}{y+1}\right) -remember, it's cyclic-

    By Cauchy-Schwarz: \text{RHS}\cdot \left(\sum (x\cdot(y+1))\right)\geq \left(\sum x \right)^2 (3)

    Let q = \sum x then : \sum (x\cdot(y+1))= q + \sum (xy) \leq q + \tfrac{q^2}{3} (*)

    Thus by (3) we have: \text{RHS}\geq \frac{q}{1+\tfrac{q}{3}}=\tfrac{3q}{3+q}

    But since q\geq 3 by AM-GM inequality -remember x\cdot y \cdot z = 1 -, it follows that \tfrac{3q}{3+q}\geq\frac{3}{2} which proves (1) \square

    (*) Since (\sum x)^2 = \sum x^2 + 2\sum xy and \sum x^2 \geq \sum xy
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