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Hard calc problem

  1. Sep 17, 2006 #1
    prove if a+b+c=1 ,a,b,c>0
    so (1+1\a)(1+1\b)91+1\c)>=64
     
  2. jcsd
  3. Sep 17, 2006 #2
    (1+1\a)(1+1\b)(1+1\c) >= 64
     
  4. Sep 18, 2006 #3

    dextercioby

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    Here's the solution

    [tex] \frac{a+1}{a}\frac{b+1}{b}\frac{c+1}{c}= \frac{abc+ab+ac+bc+2}{abc} =\frac{2}{abc}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 [/tex].

    Now, for 3 arbitrary positive real numbers the harmonic average is smaller or equal to the arithmetic average

    [tex] \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \leq \frac{a+b+c}{3}=\frac{1}{3} [/tex]

    from which it follows that

    [tex] \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 9 [/tex]

    For 3 arbitrary positive numbers, the geometric average is smaller or equal to the arithmetic average

    [tex] \sqrt[3]{abc} \leq \frac{a+b+c}{3}=\frac{1}{3} [/tex]

    from which it follows that

    [tex] \frac{1}{abc} \geq 27 [/tex].

    Now i think you easily get the wanted inequality.

    Daniel.
     
  5. Sep 19, 2006 #4
    Dexterciboy, your signature made me chuckle. It reminded me of a joke we have about one of the professors in our department, who's supposed motto we claim to be:

    "Never let an experiment get in the way of a good theory!"
     
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