14. (Poland 1) Let $P$ be any point inside a regular tetrahedron $T$. Draw 4 planes through $P$ parallel to the faces of $T$, dividing $T$ into 14 regions. From these 14 regions, remove the tetrahedra and parallelepipeds, and denote the remaining regions as $f(P)$. These regions are adjacent to edges but not to the same vertex. Given that the volume of $T$ is 1, find the minimum upper bound and the maximum lower bound of $f(P)$.

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Solution 1: Let the distances from $P$ to the four faces of $T$ be $d_{1}, d_{2}, d_{3}, d_{4}$, and the height of $T$ be $h$. Let $x_{1}=d_{1} / h$, then we have
$$
x_{1}+x_{2}+x_{3}+x_{4}=1 \text {. }
$$

Among the 14 blocks formed by the divisions, there are 4 tetrahedra, whose volumes are $x_{1}^{3}, x_{2}^{3}, x_{3}^{3}, x_{4}^{3}$; there are 4 parallelepipeds, whose volumes are $6 x_{2} x_{3} x_{4}, 6 x_{1} x_{3} x_{4}$, $6 x_{1} x_{2} x_{4}, 6 x_{1} x_{2} x_{33}$, and the sum of the volumes of the remaining blocks is
$$
\begin{aligned}
f(P) & =1-\sum_{i=1}^{4} x_{i}^{3} \\
& -6 \sum_{1 \leqslant i, j, k \leqslant 4} x_{i} x_{j} x_{k} \\
& =1-3 t+3 t^{2}+\frac{3}{4}(3 t-1) \\
& -(1-t)^{2} \\
& =\frac{3}{4}\left(3 t^{2}-3 t+1\right) t+\frac{1}{4} \geqslant \frac{1}{4} .
\end{aligned}
$$

Here, when $t=0, u=0, v=\frac{1}{4}(1-t)^{2}$, i.e., when $P$ is a point on an edge but not a vertex of $T$, the equality holds.

Clearly, when $P$ is (close to) a vertex, $f(P)$ is (close to) 0.

In summary, the upper bound and the lower bound of $f(P)$ are $\frac{3}{4}$ and 0, respectively.

Solution 2: The initial steps are the same as in Solution 1, but the specific estimation of $1-f(P)$ can be more clearly expressed.
Let $x_{1}+x_{2}=\frac{1}{2}+\delta$,
then $x_{3}+x_{4}=\frac{1}{2}-\delta$.
Here, assume $0 \leqslant \delta \leqslant \frac{1}{2}$, we get
$$
\begin{array}{l}
x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3} \\
=\left(x_{1}+x_{2}\right)^{3}+\left(x_{3}+x_{4}\right)^{3}-3\left(x_{1}^{2} x_{2}+x_{1} x_{2}^{2}+x_{3}^{2} x_{4}+x_{3} x_{4}^{2}\right) \\
=\left(\frac{1}{2}+\delta\right)^{3}+\left(\frac{1}{2}-\delta\right)^{3} \\
-3\left(\frac{1}{2}+\delta\right) x_{1} x_{2}-3\left(\frac{1}{2}-\delta\right) x_{3} x_{4} \\
=\frac{1}{4}+3 \delta^{2}-3\left[\left(\frac{1}{2}+\delta\right) x_{1} x_{2}+\left(\frac{1}{2}-\delta\right) x_{3} x_{4}\right] ; \\
6\left(x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}\right) \\
=6 x_{1} x_{2}\left(x_{3}+x_{4}\right)+6 x_{3} x_{4}\left(x_{1}+x_{2}\right) \\
=6\left(\frac{1}{2}-\delta\right) x_{1} x_{2}+6\left(\frac{1}{2}+\delta\right) x_{3} x_{4}
\end{array}
$$

Adding the two expressions, we get
$$
\begin{array}{l}
1-f(P) \\
=\frac{1}{4}+3 \delta^{2}+\left(\frac{3}{2}-9 \delta\right) x_{1} x_{2}+\left(\frac{3}{2}+9 \delta\right) x_{3} x_{4} \\
\geqslant \frac{1}{4}+3 \delta^{2}+\left(\frac{3}{2}-9 \delta\right) x_{1} x_{2} . \\
\text { If } \frac{3}{2}-9 \delta \geqslant 0, \text { it is clear that } 1-f(P) \geqslant \frac{1}{4} . \\
\text { If } \frac{3}{2}-9 \delta<0, \text { by } x_{1}+x_{2}=\frac{1}{2}+\delta \text { and the }
\end{array}
$$

arithmetic mean inequality, we have $x_{1} x_{2} \leqslant \frac{1}{4}\left(\frac{1}{2}+\delta\right)^{2}$, thus
$$
\begin{array}{l}
1-f(P) \\
\geqslant \frac{1}{4}+3 \delta^{2}+\left(\frac{3}{2}-9 \delta\right) \frac{1}{4}\left(\frac{1}{2}+\delta\right)^{2} \\
=\frac{1}{4}+\frac{3}{32}-\frac{3}{15} \delta+\frac{9}{8} \delta^{2}-\frac{9}{4} \delta^{3} \\
=\frac{1}{4}+\frac{3}{16}\left(\frac{1}{2}-\delta\right)+\frac{9}{4}\left(\frac{1}{2}-\delta\right) \delta^{2} \geqslant \frac{1}{4} .
\end{array}
$$

We still have the same result: $f(P) \leqslant \frac{3}{4}$.
When $x_{1}=1$ and $x_{2}=x_{3}=x_{4}=0$, $f(P)$
$=0$; when $x_{1}=x_{3}=\frac{1}{2}$ and $x_{2}=x_{4}=0$, $f(P)$
$=\frac{3}{4}$. Therefore, the upper bound of $f(P)$ is $\frac{3}{4}$, and the lower bound is 0.