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(C) Given $\alpha + \beta + \gamma = \pi$. For a triangle with angles $\alpha, \beta, \gamma$,the sum of sines is given by $\sin \alpha + \sin \beta +

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(C) Given $\alpha + \beta + \gamma = \pi$. For a triangle with angles $\alpha, \beta, \gamma$,the sum of sines is given by $\sin \alpha + \sin \beta + \sin \gamma = 4 \cos(\frac{\alpha}{2}) \cos(\frac{\beta}{2}) \cos(\frac{\gamma}{2})$. Since $\alpha, \beta, \gamma$ are angles of a triangle,each angle must be in the interval $(0, \pi)$,which implies $\frac{\alpha}{2}, \frac{\beta}{2}, \frac{\gamma}{2} \in (0, \frac{\pi}{2})$. In the interval $(0, \frac{\pi}{2})$,the cosine function is always positive. Therefore,$4 \cos(\frac{\alpha}{2}) \cos(\frac{\beta}{2}) \cos(\frac{\gamma}{2}) > 0$. Thus,the expression $\sin \alpha + \sin \beta + \sin \gamma$ is always positive.

$\alpha, \beta, \gamma$ are real numbers satisfying $\alpha + \beta + \gamma = \pi$. The minimum value of the expression $\sin \alpha + \sin \beta + \sin \gamma$ is

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