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Thread: absolute value

  1. #1
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    absolute value

    Challenge problem:

    Show that: $\displaystyle \frac{|a|}{|b+c|}+\frac{|b|}{|a+c|}+\frac{|c|}{|b+ a|}\geq\frac{3}{2}$,where a,b,c are any real Nos.

    $\displaystyle (|b+c|)(|a+c|)(|b+a|)\neq 0$



    Moderator edit: Approved Challenge question.
    Last edited by mr fantastic; Jun 9th 2010 at 07:43 PM. Reason: Approval.
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  2. #2
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    From $\displaystyle |x+y| \le |x| + |y|$ it follows that $\displaystyle \frac{|x|}{|y+z|} \ge \frac{|x|}{|y|+|z|}$

    Let $\displaystyle \alpha = |a|+|b|\:\:\:\:,\:\:\:\:\beta= |a|+|c|\:\:\:\:,\:\:\:\:\gamma= |b|+|c|$

    Thus, we have:

    $\displaystyle
    \frac{|a|}{|b+c|}+\frac{|b|}{|a+c|}+\frac{|c|}{|a+ b|} \ge \frac{1}{2}\left(\frac{\alpha+\beta-\gamma}{\gamma}+\frac{\alpha+\gamma-\beta}{\beta}+\frac{\gamma+\beta-\alpha}{\alpha}\right) =
    $

    $\displaystyle
    = \frac{1}{2}\left(\frac{\alpha}{\gamma}+\frac{\gamm a}{\beta}+\frac{\beta}{\alpha}\right)+\frac{1}{2}\ left(\frac{\alpha}{\beta}+\frac{\beta}{\gamma}+\fr ac{\gamma}{\alpha}\right) - \frac{3}{2} \ge
    $

    $\displaystyle
    \ge \frac{3}{2}+\frac{3}{2}-\frac{3}{2} = \frac{3}{2}
    $

    And we're done.


    On the last step we used AM-GM inequality to get

    $\displaystyle
    \frac{x}{y}+\frac{y}{z}+\frac{z}{x} \ge 3 \sqrt[3]{\frac{x}{y} \frac{y}{z} \frac{z}{x}} = 3
    $
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  3. #3
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    Quote Originally Posted by Unbeatable0 View Post
    From $\displaystyle |x+y| \le |x| + |y|$ it follows that $\displaystyle \frac{|x|}{|y+z|} \ge \frac{|x|}{|y|+|z|}$

    Let $\displaystyle \alpha = |a|+|b|\:\:\:\:,\:\:\:\:\beta= |a|+|c|\:\:\:\:,\:\:\:\:\gamma= |b|+|c|$

    Thus, we have:

    $\displaystyle
    \frac{|a|}{|b+c|}+\frac{|b|}{|a+c|}+\frac{|c|}{|a+ b|} \ge \frac{1}{2}\left(\frac{\alpha+\beta-\gamma}{\gamma}+\frac{\alpha+\gamma-\beta}{\beta}+\frac{\gamma+\beta-\alpha}{\alpha}\right) =
    $

    $\displaystyle
    = \frac{1}{2}\left(\frac{\alpha}{\gamma}+\frac{\gamm a}{\beta}+\frac{\beta}{\alpha}\right)+\frac{1}{2}\ left(\frac{\alpha}{\beta}+\frac{\beta}{\gamma}+\fr ac{\gamma}{\alpha}\right) - \frac{3}{2} \ge
    $

    $\displaystyle
    \ge \frac{3}{2}+\frac{3}{2}-\frac{3}{2} = \frac{3}{2}
    $

    And we're done.


    On the last step we used AM-GM inequality to get

    $\displaystyle
    \frac{x}{y}+\frac{y}{z}+\frac{z}{x} \ge 3 \sqrt[3]{\frac{x}{y} \frac{y}{z} \frac{z}{x}} = 3
    $
    Thanks.

    Another solution without using A.M-G.M would be if from:



    $\displaystyle
    \frac{|a|}{|b+c|}+\frac{|b|}{|a+c|}+\frac{|c|}{|a+ b|} \ge \frac{1}{2}\left(\frac{\alpha+\beta-\gamma}{\gamma}+\frac{\alpha+\gamma-\beta}{\beta}+\frac{\gamma+\beta-\alpha}{\alpha}\right) =
    $


    $\displaystyle \frac{\alpha^2+\beta^2}{2\alpha\beta}+\frac{\alpha ^2+\gamma^2}{2\alpha\gamma}+\frac{\beta^2+\gamma^2 }{2\beta\gamma}-\frac{3}{2}\geq 3-\frac{3}{2}=\frac{3}{2}$

    Since :

    $\displaystyle \alpha^2+\beta^2\geq 2\alpha\beta$ e.t.c e.t.c


    Although my own solution is completely different.
    Last edited by alexandros; Jun 10th 2010 at 06:37 AM. Reason: correction
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