To prove the given inequality, we can manipulate the right-hand side and utilize the Arithmetic Mean-Geometric Mean (AM-GM) inequality which states that for any nonnegative real numbers $x$ and $y$, we have the inequality $\frac{x + y}{2} \geq \sqrt{xy}$, with equality if and only if $x = y$.

Since $abc = 1$, we can express $a$, $b$, and $c$ as $a = \frac{1}{bc}$, $b = \frac{1}{ac}$, and $c = \frac{1}{ab}$ respectively. Let's now compare $\sqrt{a}$, $\sqrt{b}$, and $\sqrt{c}$ with their counterparts on the right side:

$$\begin{align*} \sqrt{a} + \sqrt{b} + \sqrt{c} & = \sqrt{\frac{1}{bc}} + \sqrt{\frac{1}{ac}} + \sqrt{\frac{1}{ab}} \\ & = \frac{1}{\sqrt{bc}} + \frac{1}{\sqrt{ac}} + \frac{1}{\sqrt{ab}} \\ & \frac{1}{\sqrt{bc}}, \\ \frac{1}{2} \left(\frac{1}{a} + \frac{1}{c}\right) & > \frac{1}{\sqrt{ac}}, \\ \frac{1}{2} \left(\frac{1}{a} + \frac{1}{b}\right) & > \frac{1}{\sqrt{ab}}. \end{align*}$$

Each term on the right-hand side of these inequalities corresponds to the terms on the left-hand side of our original inequality, and since $a$, $b$, and $c$ are distinct, the AM-GM inequality strictly holds (there is no equality case). Therefore, we have shown that:

$$\boxed{\sqrt{a} + \sqrt{b} + \sqrt{c} < \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}.$$