We are tasked to find the maximum value of $a + b$ under the constraint:

$\frac{1}{a^2 + 4b + 4} + \frac{1}{b^2 + 4a + 4} = \frac{1}{8}.$

Step 1: Define and simplify

Introduce the substitutions: $x = a^2 + 4b + 4, \quad y = b^2 + 4a + 4.$

Then the equation becomes: $\frac{1}{x} + \frac{1}{y} = \frac{1}{8}.$

This can be rewritten as: $\frac{x + y}{xy} = \frac{1}{8} \quad \implies \quad xy = 8(x + y).$

Step 2: Express $x$ and $y$ in terms of $a$ and $b$

From the definitions of $x$ and $y$, we have: $x = a^2 + 4b + 4, \quad y = b^2 + 4a + 4.$

Step 3: Maximization strategy

To maximize $a + b$, let us consider symmetry and equality of $a$ and $b$ (since the given equation is symmetric in $a$ and $b$).

Assume $a = b$: $x = y = a^2 + 4a + 4.$

Under this assumption, the equation becomes: $\frac{2}{x} = \frac{1}{8} \quad \implies \quad x = 16.$

Substitute $x = a^2 + 4a + 4 = 16$: $a^2 + 4a + 4 = 16 \quad \implies \quad a^2 + 4a - 12 = 0.$

Step 4: Solve for $a$

Solve the quadratic equation: $a = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm 8}{2}.$