8
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ Welcome to math.SE. For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Américo Tavares Oct 15 '12 at 21:21
  • 1
    $\begingroup$ Is the question how to prove these inequalities? If so, I'd say that explicitly. $\endgroup$ – Michael Hardy Oct 15 '12 at 21:21
  • $\begingroup$ Yes, that is the question. Thank you. $\endgroup$ – yuvalz Oct 15 '12 at 21:22
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Hint for Question2:

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

$\endgroup$
  • $\begingroup$ 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: $$2[\cos(2a)+\cos(2b)+\cos(2c)-1]\le1$$ $\endgroup$ – yuvalz Oct 16 '12 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.