Example 3 Let $P$ be a point inside (including the boundary) a regular tetrahedron $T$ with volume 1. Draw 4 planes through $P$ parallel to the 4 faces of $T$, dividing $T$ into 14 pieces. Let $f(P)$ be the sum of the volumes of those pieces which are neither tetrahedra nor parallelepipeds. Find the range of $f(P)$. (31st IMO)
Let the distances from point $P$ to the four faces of the regular tetrahedron $ABCD$ be $d_{1}, d_{2}, d_{3}, d_{4}$.
Let $x_{i}=\frac{d_{i}}{h}$, where $h$ is the height of the regular tetrahedron. Then $\sum_{i=1}^{4} x_{i}=1$.
Since the 14 pieces divided by $T$ include 4 tetrahedra with volumes $x_{i}^{3}$ each. Additionally, there are 4 parallelepipeds with volumes $6 \prod_{\substack{j \neq i \\ 1 \leqslant j \leqslant 4}} x_{j}$ each $(i=1,2,3,4)$. For example, a parallelepiped can be formed by the 3 edges starting from $A$, and by symmetry, 4 parallelepipeds can be formed. Thus,
$$f(P)=1-\sum_{i=1}^{4} x_{i}^{3}-6 \sum_{1 \leqslant i<j<k \leqslant 4} x_{i} x_{j} x_{k}$$

Clearly, $f(P) \geqslant 0$.
Furthermore, assume $x_{1}+x_{2} \leqslant \frac{1}{2}$. Let $x_{1}+x_{2}=t \leqslant \frac{1}{2}, x_{1} x_{2}=u \geqslant 0, x_{3} x_{4}=$ $v \geqslant 0$. By $\sum_{i=1}^{4} x_{i}=1$, we have
$$\begin{array}{c}
\sum_{i=1}^{4} x_{i}^{3}=\left(t^{3}-3 t u\right)+(1-t)^{3}-3(1-t) v \\
\sum_{1 \leqslant i<j<k \leqslant 4} x_{i} x_{j} x_{k}=(1-t) u+t v
\end{array}$$

Therefore,
$$\begin{aligned}
1-f(P) & =1-3 t+3 t^{2}+3(2-3 t) u+3(3 t-1) v \\
& \geqslant 1-3 t+3 t^{2}+3(3 t-1) v
\end{aligned}$$
(1) If $\frac{1}{3}<t \leqslant \frac{1}{2}$, then
$$3 t-1 \geqslant 0,1-f(P) \geqslant 1-3 t+3 t^{2} \geqslant \frac{1}{4}$$

The equality holds when $t=\frac{1}{2}, u=v=0$, i.e., when $P$ is the midpoint of an edge.
(2) If $0 \leqslant t \leqslant \frac{1}{3}$, then $3 t-1 \leqslant 0$, and $v=x_{3} x_{4} \leqslant \frac{\left(x_{3}+x_{4}\right)^{2}}{4}=\frac{(1-t)^{2}}{4}$; so
$$\begin{aligned}
1-f(P) & \geqslant 1-3 t+3 t^{2}+3(3 t-1) \cdot \frac{(1-t)^{2}}{4} \\
& =\frac{3\left(3 t^{2}+1-3 t\right) t}{4}+\frac{1}{4} \geqslant \frac{1}{4}
\end{aligned}$$

Therefore, in either case, we have $0 \leqslant f(P) \leqslant \frac{3}{4}$.
Moreover, when $P$ is a vertex of the tetrahedron, $f(P)=0$; when $P$ is the midpoint of an edge of the tetrahedron, $f(P)=$ $\frac{3}{4}$

In summary, the range of $f(P)$ is $0 \leqslant f(P) \leqslant \frac{3}{4}$.