Finding the range of $\frac{\sin(\alpha+\beta+\gamma)}{\sin\alpha+\sin\beta+\sin\gamma}$, where $\alpha,\beta,\gamma\in\left(0,\frac\pi2\right) $

trigonometry

How do I find the range of :

$$
\dfrac{\sin(\alpha +\beta +\gamma )}{\sin\alpha + \sin\beta + \sin\gamma}
$$

Where,
$$ \alpha , \beta\; and \;\gamma \in \left(0, \frac{\pi}{2}\right) $$

I tried using jensen's inequality on $\alpha, \beta \;and\; \gamma $, and got :
$$ \frac{\sin\alpha+\sin\beta+\sin\gamma}{3} \;\le\; \sin\left(\frac{\alpha + \beta+\gamma}{3}\right) $$
$$ \implies \sin\alpha+\sin\beta+\sin\gamma \;\le\; 3\sin\left(\frac{\alpha + \beta+\gamma}{3}\right) $$
Now if I assume that $\alpha+\beta+\gamma \; \le\; \pi $ then I can say:
$$ \frac{\sin\left(\alpha + \beta + \gamma\right)}{\sin\alpha+\sin\beta+\sin\gamma}\; \ge\; \frac{\sin\left(\alpha + \beta + \gamma\right)}{3\sin\left(\frac{\alpha + \beta+\gamma}{3}\right)}$$
Solving this I got:
$$ \frac{\sin\left(\alpha + \beta + \gamma\right)}{\sin\alpha+\sin\beta+\sin\gamma}\; \ge\; 1\; – \; \frac{4}{3}\sin^2\left(\frac{\alpha + \beta+\gamma}{3}\right) $$
$$\implies \frac{\sin\left(\alpha + \beta + \gamma\right)}{\sin\alpha+\sin\beta+\sin\gamma}\; > \; -\frac{1}{3}$$

But this is not useful as it is based on the assumption that $\alpha+\beta+\gamma \le \pi $.

So , I would like someone to provide me with a solution to my problem (I am not familiar with methods of higher mathematics so it would be helpful if these are not used, however use of complex numbers or any other basic inequalities in the solution is much appreciated ).

It would also be helpful if I can get more than one solution.

Thank you.

Best Answer

Using

$\displaystyle \sin(\alpha+\beta+\gamma)=\sin\alpha\cos\beta\cos\gamma+\cos\alpha\sin\beta\cos\gamma+\cos\alpha\cos\beta\sin\gamma-\sin\alpha\sin\beta\sin\gamma$

$\displaystyle \Longrightarrow \sin(\alpha+\beta+\gamma)-\sin\alpha-\sin\beta-\sin\gamma=\sin\alpha(\underbrace{\cos\beta\cos\gamma-1}_{<0})+\sin\beta(\underbrace{\cos\alpha\cos\gamma-1}_{<0})+\sin\gamma(\underbrace{\cos\alpha\cos\beta-1}_{<0})-\sin\alpha\sin\beta\sin\gamma<0$

$\displaystyle \sin(\alpha+\beta+\gamma)<\sin\alpha+\sin\beta+\sin\gamma$

$\displaystyle \frac{\sin(\alpha+\beta+\gamma)}{\sin\alpha+\sin\beta+\sin\gamma}<1$

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