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Tough Olympiad-like Inequalities question

  1. Apr 6, 2006 #1
    a, b, c, and d are all positive real numbers.

    Given that

    a + b + c + d = 12
    abcd = 27 + ab +ac +ad + bc + bd + cd

    Determine a, b, c, and d.

    ---

    The solution says that using AM - GM on the second equation gives

    abcd (is greater than or equal to) 27 + 6*sqrt of (abcd)

    From there they rewrite the second equation as:

    abcd - 6sqrt(abcd) - 27 (is greater than or equal to) 0, resulting in:

    sqrt(abcd) (is greater than or equal to) 9
    and thus abcd ^ (1/4), which is the GM, is greater than or equal to 3

    But according to AM - GM, the AM of a, b, c, and d, which is equal to 3, is greater than or equal to the GM, (abcd) ^ (1/4)

    Therefore:

    3 (is greater than or equal to) abcd ^ (1/4), but from previous, the GM is greater than or equal to 3. This can only occur if the AM and GM are both 3.

    *The only step that I don't understand is how they applied AM - GM to

    abcd = 27 + ab +ac +ad + bc + bd + cd

    to obtain

    abcd (is greater than or equal to) 27 + 6*sqrt of (abcd)

    Any help would be appreciated! Thank you
     
  2. jcsd
  3. Apr 7, 2006 #2

    VietDao29

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    Homework Helper

    Hmm, you should note that:
    [tex]ab + cd \geq 2 \sqrt{abcd}[/tex]
    [tex]ac + bd \geq 2 \sqrt{abcd}[/tex]
    [tex]ad + bc \geq 2 \sqrt{abcd}[/tex]
    Adding both sides of the 4 inequalities, we have:
    [tex]ab +ac +ad + bc + bd + cd \geq 6 \sqrt{abcd}[/tex]
    So that means:
    [tex]abcd = 27 + ab +ac +ad + bc + bd + cd \geq 27 + 6\sqrt{abcd}[/tex]
    Can you get this? :)
     
  4. Apr 7, 2006 #3

    0rthodontist

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    Science Advisor

    I got it in 1 minute by guessing.
     
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