Trigonometric Inequality

**great_math** · June 7th 2009, 11:02 AM

If $\alpha,\beta$ and $\gamma$ be the angles in any acute triangle. Prove that

$2<\sin \alpha + \sin \beta + \sin \gamma \le \dfrac{3\sqrt{3}}{2}$

**alexmahone** · June 7th 2009, 08:00 PM

Using Jensen's inequality,

$\frac{sin \alpha+sin \beta+sin \gamma}{3} \leq sin (\frac{\alpha + \beta + \gamma}{3})$
$\frac{sin \alpha+sin \beta+sin \gamma}{3} \leq sin (\frac{\pi}{3})$
$\frac{sin \alpha+sin \beta+sin \gamma}{3} \leq \frac{\sqrt3}{2}$
$\sin \alpha + \sin \beta + \sin \gamma \le \dfrac{3\sqrt{3}}{2}$

**Rachel.F** · June 8th 2009, 02:58 AM

**Isomorphism** · June 10th 2009, 03:21 AM

Originally Posted by Rachel.F

Thats wrong!

$\sin \beta < 1, \sin \gamma < 1$ does not imply the last step in your proof. This part of the question was posted by fardeen_gen recently.

**Rachel.F** · June 10th 2009, 03:27 AM

Originally Posted by Isomorphism

Thats wrong!

$\sin \beta < 1, \sin \gamma < 1$ does not imply the last step in your proof. This part of the question was posted by fardeen_gen recently.

why? i didn't understand

**Isomorphism** · June 10th 2009, 03:38 AM

Originally Posted by Isomorphism

Thats wrong!

$\sin \beta < 1, \sin \gamma < 1$ does not imply the last step in your proof. This part of the question was posted by fardeen_gen recently.

Originally Posted by Rachel.F

why? i didn't understand

$\sin \beta < 1, \sin \gamma < 1 \implies \sin \beta + \sin \gamma < 2$

Now $\sin \alpha + \sin \beta + \sin \gamma > \sin \beta + \sin \gamma$ , however the right hand side of the inequality is less than 2. This says nothing about the sum of the sines.

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