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This is the first time I'm posting here. If you can also tell me how to format this like a pro, I'll be very grateful.

1st question:

Prove the following inequality: $0^{\circ} < a, b, c < 180^{\circ}$

$\sin a \times \sin b \times \sin c \le \sin\left(\frac{a+b}{2}\right) \times \sin\left(\frac{a+c}{2}\right) \times \sin\left(\frac{a+b}{2}\right)$

2nd question:

Prove the following inequality: $a + b + c = 90^{\circ}$

$\sin a \times \sin b \times \sin c \le \frac{1}{8}$

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    Yes, that is the question. Thank you.2012-10-15

2 Answers 2

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First inequality By Jensen we have

$\sin\left(\frac{a+b}{2}\right) \geq \frac{\sin(a)+\sin(b)}{2}$

By AM-GM we get

$\frac{\sin(a)+\sin(b)}{2} \geq \sqrt{ \sin(a) \sin(b)}$

Combining we get

$\sqrt{\sin(a) \sin(b) } \leq \sin\left(\frac{a+b}{2}\right)$

Similarly you get

$\sqrt{\sin(a) \sin(c) } \leq \sin\left(\frac{a+c}{2}\right)$

$\sqrt{\sin(b) \sin(c) } \leq \sin\left(\frac{b+c}{2}\right)$

Multiplying them you get the desired inequality.

Second Inequality:

By AM-GM:

$\sin a \times \sin b \times \sin c \le \left( \frac{\sin(a)+\sin(b)+\sin(c)}{3} \right)^3$

Now, by Jensen:

$\frac{\sin(a)+\sin(b)+\sin(c)}{3} \leq \sin(\frac{a+b+c}{3})=\frac{1}{2}$

Combining the two Yields the desired result.

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Hint for Question2:

$\sin{a}\sin{b}\sin{c}=\frac{1}{2}[\cos(a-b)\sin{c}-\cos(a+b)\sin{c}]=\frac{1}{4}[\sin(a-b+c)+\sin(c-a+b)-\sin(a+b+c)-\sin(c-a-b)$

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    First of all, thank you. I reached that point myself, but don't know how to continue. I also reached this point for the 1st question: $\sin(a+b-c)+\sin(a+c-b)+\sin(b+c-a)\le\sin(a)+\sin(b)+\sin(c)$ ---------- And this point for the 2nd question: $2[\cos(2a)+\cos(2b)+\cos(2c)-1]\le1$2012-10-16