1. Let $A', B', C'$ be the projections of a point $M$ inside a triangle $ABC$ onto the sides $BC, CA, AB$, respectively. Define $MA' = d_a$, $MB' = d_b$, $MC' = d_c$, $\angle MAB = \alpha_c$, and $\angle MAC = \alpha_b$.

2. We need to find the position of point $M$ that maximizes $p(M) = \frac{MA' \cdot MB' \cdot MC'}{MA \cdot MB \cdot MC}$.

3. Consider the product $MB \cdot MC$. Using the Law of Sines in triangle $MBC$, we have:
$$MB \cdot MC = \frac{d_b d_c}{\sin \alpha_c \sin \alpha_b}$$

4. Using the trigonometric identity for the product of sines, we get:
$$\sin \alpha_c \sin \alpha_b = \frac{1}{2} [\cos(\alpha_c - \alpha_b) - \cos(\alpha_c + \alpha_b)]$$

5. Since $\cos(\alpha_c + \alpha_b) \leq 1$, we have:
$$\cos(\alpha_c - \alpha_b) - \cos(\alpha_c + \alpha_b) \geq \cos(\alpha_c - \alpha_b) - 1$$

6. Therefore:
$$MB \cdot MC \geq \frac{2 d_b d_c}{1 - \cos A} = \frac{d_b d_c}{\sin^2 \frac{A}{2}}$$

7. Similarly, we can derive:
$$MC \cdot MA \geq \frac{d_c d_a}{\sin^2 \frac{B}{2}}$$
$$MA \cdot MB \geq \frac{d_a d_b}{\sin^2 \frac{C}{2}}$$

8. Multiplying these inequalities together and taking the square root, we get:
$$\frac{d_a d_b d_c}{MA \cdot MB \cdot MC} \leq \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$$

9. Equality holds if and only if $M \equiv I$, the incenter of $\triangle ABC$.

The final answer is $\boxed{ M \equiv I }$, the incenter of $\triangle ABC$.