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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: $$a + b + c = 90^{\circ}$$

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

yuvalz
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2 Answers2

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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.

N. S.
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Hint for Question2:

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

Salech Alhasov
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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: $$2[\cos(2a)+\cos(2b)+\cos(2c)-1]\le1$$

     – yuvalz Oct 16 '12 at 14:31