4. Let $M$ be a point inside $\triangle ABC$, and let $M A^{\prime}$ be perpendicular to $BC$, with the foot of the perpendicular being $A^{\prime}$. Similarly, define $B^{\{\prime\}}$ on $CA$ and $C^{\prime}$ on $AB$. Let
$$
P(M)=\frac{M A^{\prime} \cdot M B^{\prime} \cdot M C^{\prime}}{M A \cdot M B \cdot M C} .
$$

Determine the position of $M$ such that $P(M)$ is maximized. Let $\mu(ABC)$ denote this maximum value. What condition must the triangle satisfy for $\mu(ABC)$ to be maximized?
Let $\angle M A B=\alpha_{1}, \angle M A C=\alpha_{2}, \angle M B C=\beta_{1}$, $\angle M B A=\beta_{2}, \angle M C A=\gamma_{1}, \angle M C B=\gamma_{2}$. Then
$$
\begin{array}{l}
\frac{M B^{\prime} \cdot M C^{\prime}}{M A^{2}}=\sin \alpha_{1} \cdot \sin \alpha_{2} \\
=\frac{1}{2}\left[\cos \left(\alpha_{1}-\alpha_{2}\right)-\cos \left(\alpha_{1}+\alpha_{2}\right)\right] \\
\leqslant \frac{1}{2}(1-\cos A)=\sin ^{2} \frac{A}{2} .
\end{array}
$$

Similarly, $\frac{M 4^{\prime} \cdot M C^{\prime}}{M B^{2}}=\sin \beta_{1} \cdot \sin \beta_{2} \leqslant \sin ^{2} \frac{B}{2}$,
$$
\frac{M B^{\prime} \cdot M A^{\prime}}{M C^{2}}=\sin \gamma_{1} \cdot \sin \gamma_{2} \leqslant \sin ^{2} \frac{C}{2} \text {. }
$$

Then $P^{2}(M)=\sin \alpha_{1} \cdot \sin \alpha_{2} \cdot \sin \beta_{1} \cdot \sin \beta_{2} \cdot \sin \gamma_{1} \cdot \sin \gamma_{2}$
$$
\leqslant \sin ^{2} \frac{A}{2} \cdot \sin ^{2} \frac{B}{2} \cdot \sin ^{2} \frac{C}{2} .
$$

Equality holds if and only if $\alpha_{1}=\alpha_{2}, \beta_{1}=\beta_{2}, \gamma_{1}=\gamma_{2}$, i.e., when $M$ is the incenter of $\triangle A B C$, $P(M)$ achieves its maximum value, and the maximum value is
$$
\mu(A B C)=\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} .
$$

By the AM-GM inequality and Jensen's inequality, we have
$$
\begin{array}{l}
\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} \leqslant\left(\frac{\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}}{3}\right)^{3} \\
\leqslant\left(\sin \frac{\frac{A}{2}+\frac{B}{2}+\frac{C}{2}}{3}\right)^{3}=\sin ^{3} \frac{\pi}{6}=\frac{1}{8} .
\end{array}
$$

Equality holds if and only if $\triangle A B C$ is an equilateral triangle.