4. As shown in Figure 2, the incircle of $\triangle A B C$ touches sides $A B$, $A C$, and $B C$ at points $E$, $F$, and $G$ respectively. If the product of the lengths of the three sides of $\triangle A B C$ is 64, then the maximum value of $A E \cdot B G \cdot C F$ is ( ).
(A) 6
(B) 8
(C) 10
(D) 16
4. B.

Let $BC=a$, $AC=b$, $AB=c$. Then $abc=64$, $AE \cdot BG \cdot CF=\frac{a+b-c}{2} \cdot \frac{a+c-b}{2} \cdot \frac{b+c-a}{2}$.
Thus, we only need to prove:
$$
(a+b-c)(a+c-b)(b+c-a) \leqslant abc \text {. }
$$

Let $a+b-c=x$, $a+c-b=y$, $b+c-a=z$.
Then equation (1) becomes $xyz \leqslant \frac{x+y}{2} \cdot \frac{x+z}{2} \cdot \frac{y+z}{2}$.
The above inequality can be derived from $\frac{x+y}{2} \geqslant \sqrt{xy}$.