- #1

- 4

- 0

(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?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter endangered
- Start date

- #1

- 4

- 0

(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

CompuChip

Science Advisor

Homework Helper

- 4,302

- 47

[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]

Share: