1. Substitution: Let $a = 2x$, $b = 2y$, and $c = 2z$. Given $abc = 8$, we have:
$$(2x)(2y)(2z) = 8 \implies 8xyz = 8 \implies xyz = 1$$
We need to prove:
$$\frac{ab + 4}{a + 2} + \frac{bc + 4}{b + 2} + \frac{ca + 4}{c + 2} \geq 6$$

2. Simplification: Substitute $a = 2x$, $b = 2y$, and $c = 2z$ into the expression:
$$\frac{ab + 4}{a + 2} = \frac{(2x)(2y) + 4}{2x + 2} = \frac{4xy + 4}{2(x + 1)} = \frac{2(xy + 1)}{x + 1}$$
Similarly,
$$\frac{bc + 4}{b + 2} = \frac{2(yz + 1)}{y + 1}$$
$$\frac{ca + 4}{c + 2} = \frac{2(zx + 1)}{z + 1}$$
Therefore, the inequality becomes:
$$\frac{2(xy + 1)}{x + 1} + \frac{2(yz + 1)}{y + 1} + \frac{2(zx + 1)}{z + 1} \geq 6$$
Dividing both sides by 2, we need to prove:
$$\frac{xy + 1}{x + 1} + \frac{yz + 1}{y + 1} + \frac{zx + 1}{z + 1} \geq 3$$

3. Further Substitution: Let $x = \frac{p}{q}$, $y = \frac{q}{r}$, and $z = \frac{r}{p}$. Then $xyz = 1$ is satisfied because:
$$\left(\frac{p}{q}\right)\left(\frac{q}{r}\right)\left(\frac{r}{p}\right) = 1$$
We need to prove:
$$\sum_{cyc} \frac{xy + 1}{x + 1} = \frac{\frac{p}{q} \cdot \frac{q}{r} + 1}{\frac{p}{q} + 1} = \frac{\frac{p}{r} + 1}{\frac{p + q}{q}} = \frac{\frac{p + r}{r}}{\frac{p + q}{q}} = \frac{q(p + r)}{r(p + q)} \geq 3$$

4. Application of AM-GM Inequality: By the AM-GM inequality, we know that for positive real numbers $a, b, c$:
$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3$$
Applying this to our terms:
$$\frac{q(p + r)}{r(p + q)} + \frac{r(q + p)}{p(q + r)} + \frac{p(r + q)}{q(r + p)} \geq 3$$
This is straightforward by AM-GM since the product of all summands is 1.

Thus, we have shown that:
$$\frac{xy + 1}{x + 1} + \frac{yz + 1}{y + 1} + \frac{zx + 1}{z + 1} \geq 3$$