Prove that if the area of the face opposite vertex $B_{i}(i=1,2,3,4)$ is denoted as $s_{i}(i=1,2,3,4)$, and in equation (8) we take $x_{i}=\frac{\left|P C_{i}\right|}{s_{i}}(i=1,2,3,4)$, then the left side of equation (8) is
$$\begin{array}{l} x_{1} s_{1}^{2}+x_{2} s_{2}^{2}+x_{3} s_{3}^{2}+x_{4} s_{4}^{2}= \\ \left|P C_{1}\right| \cdot s_{1}+\left|P C_{2}\right| \cdot s_{2}+\left|P C_{3}\right| \cdot s_{3}+\left|P C_{4}\right| \cdot s_{4}= \end{array}$$
$$3 B$$

Furthermore, if the dihedral angles of the tetrahedron $P-C_{1} C_{2} C_{3}$ are $\angle C_{2} P C_{3}=\alpha_{1}, \angle C_{3} P C_{1}=\alpha_{2}, \angle C_{1} P C_{2}=\alpha_{3}$, then the dihedral angles formed by the edges $B_{4} B_{1}, B_{4} B_{2}, B_{4} B_{3}$ are complementary to $\alpha_{1}, \alpha_{2}, \alpha_{3}$. According to the volume formula of a tetrahedron (proof omitted), we have
$$\begin{array}{l} V_{P-C_{1} C_{2} c_{3}}=\frac{1}{6}\left|P C_{1}\right|\left|P C_{2}\right|\left|P C_{3}\right| \cdot \\ \sqrt{1-\cos ^{2} \alpha_{1}-\cos ^{2} \alpha_{2}-\cos ^{2} \alpha_{3}+2 \cos \alpha_{1} \cos \alpha_{2} \cos \alpha_{3}} \\ B^{2}=\left(V_{B_{4}-B_{1} B_{2} B_{3}}\right)^{2}=\frac{2}{9} s_{1} s_{2} s_{3} \cdot \\ \sqrt{1-\cos ^{2}\left(\pi-\alpha_{1}\right)-\cos ^{2}\left(\pi-\alpha_{2}\right)-\cos ^{2}\left(\pi-\alpha_{3}\right)-2 \cos \left(\pi-\alpha_{1}\right) \cos \left(\pi-\alpha_{2}\right) \cos \left(\pi-\alpha_{3}\right)}= \\ \frac{2}{9} s_{1} s_{2} s_{3} \sqrt{1-\cos ^{2} \alpha_{1}-\cos ^{2} \alpha_{2}-\cos ^{2} \alpha_{3}+2 \cos \alpha_{1} \cos \alpha_{2} \cos \alpha_{3}} \\ \quad \frac{V_{P-C_{1} C_{2} C_{3}}}{B^{2}}=\frac{3\left|P C_{1}\right|\left|P C_{2}\right|\left|P C_{3}\right|}{4 s_{1} s_{2} s_{3}} \\ \text { Therefore, } \end{array}$$
$$x_{1} x_{2} x_{3}=\frac{\left|P C_{1}\right|}{s_{1}} \cdot \frac{\left|P C_{2}\right|}{s_{2}} \cdot \frac{\left|P C_{3}\right|}{s_{3}}=\frac{4 V_{P-c_{1} c_{2} C_{3}}}{3 B^{2}}$$

Similarly, we have $x_{2} x_{3} x_{4}=\frac{4 V_{P-c_{2} c_{3} c_{4}}}{3 B^{2}}, x_{1} x_{3} x_{4}=\frac{4 V_{P-c_{1} c_{3} c_{4}}}{3 B^{2}}, x_{1} x_{2} x_{4}=\frac{4 V_{P-c_{1} c_{2} c_{4}}}{3 B^{2}}$, thus we can obtain
$$\begin{array}{l} x_{2} x_{3} x_{4}+x_{1} x_{3} x_{4}+x_{1} x_{2} x_{4}+x_{1} x_{2} x_{3}= \\ \frac{4}{3}\left(\frac{V_{P-c_{2} c_{3} c_{4}}}{B^{2}}+\frac{V_{P-c_{1} c_{3} c_{4}}}{B^{2}}+\frac{V_{P-c_{1} c_{2} c_{4}}}{B^{2}}+\frac{V_{P-c_{1} c_{2} c_{3}}}{B^{2}}\right)= \\ \frac{4 C}{3 B^{2}} \end{array}$$

Thus, the right side of equation (8) is
$$\begin{array}{l} \frac{9 \sqrt[3]{6}}{2}\left(x_{2} x_{3} x_{4}+x_{1} x_{3} x_{4}+x_{1} x_{2} x_{4}+x_{1} x_{2} x_{3}\right)^{\frac{1}{3}} \cdot B^{\frac{4}{3}}= \\ \frac{9 \sqrt[3]{6}}{2}\left(\frac{4 C}{3 B^{2}}\right)^{\frac{1}{3}} \cdot B^{\frac{4}{3}}= \\ 9 B^{\frac{2}{3}} \cdot C^{\frac{1}{3}} \end{array}$$

According to the inequality (8), we get
$$\begin{aligned} 3 B & \geqslant 9 B^{\frac{2}{3}} \cdot C^{\frac{1}{3}} \\ C & \leqslant \frac{1}{27} B \end{aligned}$$

Proof complete. $\square$