The minimum value of $$\alpha$$ for which the equation $$\frac{4}{\sin x} + \frac{1}{1 - \sin x} = \alpha$$ has at least one solution in $$\left(0, \frac{\pi}{2}\right)$$ is ______.

We need to find the minimum value of $$\alpha$$ for which the equation $$\frac{4}{\sin x} + \frac{1}{1 - \sin x} = \alpha$$ has at least one solution in $$\left(0, \frac{\pi}{2}\right)$$.

Let $$t = \sin x$$. Since $$x \in \left(0, \frac{\pi}{2}\right)$$, we have $$t \in (0, 1)$$.

We define $$f(t) = \frac{4}{t} + \frac{1}{1 - t}$$ for $$t \in (0, 1)$$. The equation has a solution if and only if $$\alpha$$ is in the range of $$f(t)$$.

To find the minimum of $$f(t)$$, we compute the derivative: $$f'(t) = -\frac{4}{t^2} + \frac{1}{(1-t)^2}$$.

Setting $$f'(t) = 0$$, we get $$\frac{1}{(1-t)^2} = \frac{4}{t^2}$$, so $$t^2 = 4(1-t)^2$$.

Taking the positive square root (since $$t > 0$$ and $$1 - t > 0$$), $$t = 2(1 - t)$$, which gives $$t = 2 - 2t$$, so $$3t = 2$$ and $$t = \frac{2}{3}$$.

Now $$f\left(\frac{2}{3}\right) = \frac{4}{2/3} + \frac{1}{1 - 2/3} = 6 + 3 = 9$$.

We verify this is a minimum by checking the second derivative or noting that $$f(t) \to \infty$$ as $$t \to 0^+$$ and $$t \to 1^-$$, so the critical point must be a minimum.

Hence, the answer is $$9$$.