Arithmetic mean always greater than geometric mean

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SUMMARY

The discussion centers on the mathematical inequality stating that the arithmetic mean of sine values, specifically (sin A + sin B + sin C)/3, is always greater than or equal to the geometric mean, represented as √[3]{sin A * sin B * sin C}. The user seeks alternative proofs for this inequality, suggesting a potential approach using cubic means. The conversation highlights the established relationship between arithmetic and geometric means in the context of trigonometric functions.

PREREQUISITES
  • Understanding of trigonometric functions, particularly sine.
  • Familiarity with mathematical inequalities, specifically the Arithmetic Mean-Geometric Mean (AM-GM) inequality.
  • Knowledge of cubic means and their properties.
  • Basic algebraic manipulation skills for inequalities.
NEXT STEPS
  • Study the proof of the Arithmetic Mean-Geometric Mean inequality in detail.
  • Explore the properties of cubic means and their applications in inequalities.
  • Investigate alternative proofs for inequalities involving trigonometric functions.
  • Learn about Jensen's inequality and its relevance to convex functions.
USEFUL FOR

Mathematicians, students studying inequalities, and anyone interested in advanced trigonometric properties and proofs.

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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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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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