$23 \cdot 14$ Let the areas of the four faces of a tetrahedron be $S_{1}, S_{2}, S_{3}, S_{4}$, and their maximum value be $S$, and let $\lambda=\frac{\sum_{i=1}^{4} S_{i}}{S}$, then $\lambda$ must satisfy
(A) $2<\lambda \leqslant 4$.
(B) $3<\lambda<4$.
(C) $2.5<\lambda \leqslant 4.5$.
(D) $3.5<\lambda<5.5$.
(China High School Mathematics League, 1992)
$$
\begin{array}{l}
\text { [Solution] } \because S_{i} \leqslant S(i=1,2,3,4), \\
\therefore \quad \sum_{i=1}^{4} S_{i} \leqslant 4 S .
\end{array}
$$

Thus, $\quad \frac{\sum_{i=1}^{4} S_{i}}{S} \leqslant 4$.
In particular, when the tetrahedron is a regular tetrahedron, the equality in the above equation holds, thus negating (B).

Consider any regular triangular pyramid where the dihedral angle between the lateral face and the base is $45^{\circ}$, and let $S_{4}$ represent the area of its base, then
$$
S=S_{4}=\left(S_{1}+S_{2}+S_{3}\right) \cos 45^{\circ}=\frac{\sqrt{2}}{2}\left(S_{1}+S_{2}+S_{3}\right),
$$

Thus,
$$
\begin{array}{l}
S_{1}+S_{2}+S_{3}=\sqrt{2} S \\
S_{1}+S_{2}+S_{3}+S_{4}=(1+\sqrt{2}) S
\end{array}
$$

That is,
then $2<\lambda=\frac{\sum_{i=1}^{4} S_{i}}{S}=1+\sqrt{2}<2.5$, thus negating $(C),(D)$. Therefore, the correct choice is $(A)$.