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Math Help - absolute value

  1. #1
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    absolute value

    Challenge problem:

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

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



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

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

    Thus, we have:

    <br />
\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) =<br />

    <br />
 = \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<br />

    <br />
\ge \frac{3}{2}+\frac{3}{2}-\frac{3}{2} = \frac{3}{2}<br />

    And we're done.


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

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

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

    Thus, we have:

    <br />
\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) =<br />

    <br />
 = \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<br />

    <br />
\ge \frac{3}{2}+\frac{3}{2}-\frac{3}{2} = \frac{3}{2}<br />

    And we're done.


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

    <br />
\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \ge 3 \sqrt[3]{\frac{x}{y} \frac{y}{z} \frac{z}{x}} = 3<br />
    Thanks.

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



    <br />
\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)  =<br />


    \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 :

    \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; June 10th 2010 at 06:37 AM. Reason: correction
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