# Math Help - Trigonometric Inequality

1. ## Trigonometric Inequality

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

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

2. Using Jensen's inequality,

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

3. 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.

4. 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

5. 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.