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Math Help - proving for real positive numbers

  1. #1
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    Post proving for real positive numbers

    how can i do?
    For a,b,c real positive numbers, show that ;

    (a+b+c)(1/a + 1/b + 1/c)>= 9

    is there any theorem to use?
    i dont know theorems based on this..
    can u help me with this?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Using the AM-GM-HM inequality,

    AM = \frac{a+b+c}{3}

    HM = \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}

    AM \geq GM \geq HM

    AM \geq HM

    \frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}

    (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq 9
    Last edited by alexmahone; July 23rd 2009 at 12:57 AM.
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  3. #3
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    Thumbs up thank u very much..

    i don't even know the fact that .
    thankx for ur help alexmahone.. i really appreciate it..
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  4. #4
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    You could start with
     (a - b )^2 \geq 0, \mbox{ so } a^2 + b^2 - 2ab \geq 0 .
    Then,
    \ a^2 + b^2 \geq 2ab \mbox{ so } \frac{a^2 + b^2}{ab} \geq 2 \mbox{ ( given that } a \mbox{ and } b are positive numbers).
    So, multiply out the LHS of your inequality and collect up terms. The two terms involving  a \mbox{ and } b can be combined to produce the fraction  ( a^2 + b^2 ) / ab , so you can make use of the result obtained earlier.
    Do the same for the terms involving  b \mbox{ and } c and those involving  c \mbox{ and } a , and the result follows.
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