Find positive reals $a, b, c$ which maximizes the value of $abc$ subject to the constraint that $b(a^2 + 2) + c(a + 2) = 12$.
To solve the problem, we need to maximize the value of \(abc\) subject to the constraint \(b(a^2 + 2) + c(a + 2) = 12\). We will use the method of Lagrange multipliers to find the maximum value.

1. **Define the function and constraint:**
   Let \(f(a, b, c) = abc\) be the function we want to maximize, and let \(g(a, b, c) = b(a^2 + 2) + c(a + 2) - 12\) be the constraint.

2. **Set up the Lagrange function:**
   The Lagrange function is given by:
   \[
   \mathcal{L}(a, b, c, \lambda) = abc + \lambda (12 - b(a^2 + 2) - c(a + 2))
   \]

3. **Compute the partial derivatives:**
   \[
   \frac{\partial \mathcal{L}}{\partial a} = bc - \lambda (2ab + c) = 0
   \]
   \[
   \frac{\partial \mathcal{L}}{\partial b} = ac - \lambda (a^2 + 2) = 0
   \]
   \[
   \frac{\partial \mathcal{L}}{\partial c} = ab - \lambda (a + 2) = 0
   \]
   \[
   \frac{\partial \mathcal{L}}{\partial \lambda} = 12 - b(a^2 + 2) - c(a + 2) = 0
   \]

4. **Solve the system of equations:**
   From \(\frac{\partial \mathcal{L}}{\partial b} = 0\):
   \[
   ac = \lambda (a^2 + 2) \implies \lambda = \frac{ac}{a^2 + 2}
   \]

   From \(\frac{\partial \mathcal{L}}{\partial c} = 0\):
   \[
   ab = \lambda (a + 2) \implies \lambda = \frac{ab}{a + 2}
   \]

   Equating the two expressions for \(\lambda\):
   \[
   \frac{ac}{a^2 + 2} = \frac{ab}{a + 2}
   \]

   Cross-multiplying gives:
   \[
   ac(a + 2) = ab(a^2 + 2)
   \]
   \[
   aca + 2ac = aba^2 + 2ab
   \]
   \[
   ac(a + 2) = ab(a^2 + 2)
   \]
   \[
   c(a + 2) = b(a^2 + 2)
   \]

5. **Substitute back into the constraint:**
   Using \(c(a + 2) = b(a^2 + 2)\), we substitute into the constraint:
   \[
   b(a^2 + 2) + c(a + 2) = 12
   \]
   \[
   b(a^2 + 2) + b(a^2 + 2) = 12
   \]
   \[
   2b(a^2 + 2) = 12
   \]
   \[
   b(a^2 + 2) = 6
   \]
   \[
   b = \frac{6}{a^2 + 2}
   \]

6. **Find \(c\):**
   Using \(c(a + 2) = b(a^2 + 2)\):
   \[
   c(a + 2) = \frac{6}{a^2 + 2}(a^2 + 2) = 6
   \]
   \[
   c = \frac{6}{a + 2}
   \]

7. **Calculate \(abc\):**
   \[
   abc = a \cdot \frac{6}{a^2 + 2} \cdot \frac{6}{a + 2}
   \]
   \[
   abc = \frac{36a}{(a^2 + 2)(a + 2)}
   \]

8. **Maximize \(abc\):**
   To maximize \(abc\), we need to find the critical points of \(\frac{36a}{(a^2 + 2)(a + 2)}\). This can be done by taking the derivative with respect to \(a\) and setting it to zero. However, we can also use the AM-GM inequality to find an upper bound.

   By AM-GM inequality:
   \[
   a^2 + 2 \geq 2\sqrt{2a^2} = 2a
   \]
   \[
   a + 2 \geq 2\sqrt{a}
   \]

   Therefore:
   \[
   (a^2 + 2)(a + 2) \geq 4a
   \]

   Hence:
   \[
   abc = \frac{36a}{(a^2 + 2)(a + 2)} \leq \frac{36a}{4a} = 9
   \]

   However, we need to check if this bound is achievable. By setting \(a = 2\), we get:
   \[
   b = \frac{6}{2^2 + 2} = \frac{6}{6} = 1
   \]
   \[
   c = \frac{6}{2 + 2} = \frac{6}{4} = \frac{3}{2}
   \]

   Then:
   \[
   abc = 2 \cdot 1 \cdot \frac{3}{2} = 3
   \]

   This shows that the maximum value of \(abc\) is indeed \(\boxed{3}\).