Arithmetic mean always greater than geometric mean

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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?
 

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  • #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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