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Arithmetic mean always greater than geometric mean

  1. May 17, 2008 #1
    Hey,

    (sin A + sin B + sin C)/3 >= [tex]\sqrt[3]{}[/tex](sin A*sin B*sin C)

    I know this is true by Arithmetic mean always greater than geometric mean...
    but is there any other way of proving this?
     
  2. jcsd
  3. May 17, 2008 #2

    CompuChip

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    What about something along the lines of:

    [tex]\left[ \frac13 (\sin A + \sin B + \sin C) \right]^3 \ge \frac19 \left( \sin^3 A + \sin^3 B + \sin^3 C \right) \ge \frac13 \sin^3 A \ge \sin A \sin B \sin C[/tex]
     
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