If $\displaystyle \alpha,\beta$ and $\displaystyle \gamma$ be the angles in any acute triangle. Prove that
$\displaystyle 2<\sin \alpha + \sin \beta + \sin \gamma \le \dfrac{3\sqrt{3}}{2}$
If $\displaystyle \alpha,\beta$ and $\displaystyle \gamma$ be the angles in any acute triangle. Prove that
$\displaystyle 2<\sin \alpha + \sin \beta + \sin \gamma \le \dfrac{3\sqrt{3}}{2}$
Using Jensen's inequality,
$\displaystyle \frac{sin \alpha+sin \beta+sin \gamma}{3} \leq sin (\frac{\alpha + \beta + \gamma}{3})$
$\displaystyle \frac{sin \alpha+sin \beta+sin \gamma}{3} \leq sin (\frac{\pi}{3})$
$\displaystyle \frac{sin \alpha+sin \beta+sin \gamma}{3} \leq \frac{\sqrt3}{2}$
$\displaystyle \sin \alpha + \sin \beta + \sin \gamma \le \dfrac{3\sqrt{3}}{2}$
$\displaystyle \sin \beta < 1, \sin \gamma < 1 \implies \sin \beta + \sin \gamma < 2 $
Now $\displaystyle \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.