Problem 1262

Proof (I) Let $O, G$ be the circumcenter and centroid of tetrahedron $A_{1} A_{2} A_{3} A_{4}$, respectively. Extend $A_{1} G$ to intersect the plane $A_{2} A_{3} A_{4}$ at $G_{1}$, then $G_{1}$ is the centroid of $\triangle A_{2} A_{3} A_{4}$. Connect $A_{2} G_{1}$ to intersect $A_{3} A_{4}$ at point $M$, then $M$ is the midpoint of $A_{3} A_{4}$, as shown in Figure 22-9.
From the proof in Example 9, we know that
$$A_{1} G_{1}^{2}=\frac{1}{9}\left[3\left(A_{1} A_{2}^{2}+A_{1} A_{3}^{2}+A_{1} A_{4}^{2}\right)-\left(A_{2} A_{3}^{2}+A_{2} A_{4}^{2}+A_{3} A_{4}^{2}\right)\right].$$
Similarly, in the tetrahedron $O A_{2} A_{3} A_{4}$, we have
$$\begin{aligned} O G_{1}^{2} & =\frac{1}{9}\left[3\left(O A_{2}^{2}+O A_{3}^{2}+O A_{4}^{2}\right)-\left(A_{2} A_{3}^{2}+A_{2} A_{4}^{2}+A_{3} A_{4}^{2}\right)\right] \\ & =R^{2}-\frac{1}{9}\left(A_{2} A_{3}^{2}+A_{2} A_{4}^{2}+A_{3} A_{4}^{2}\right). \end{aligned}$$

Since $G$ is the centroid of the tetrahedron, by property 3, we know $\frac{A_{1} G}{G G_{1}}=\frac{3}{1}$. Thus, in $\triangle A_{1} O G_{1}$, applying Stewart's theorem to point $G$, we have
$$\begin{aligned} O G^{2} & =\frac{1}{16}\left[4\left(3 O G_{1}^{2}+O A_{1}^{2}\right)-3 A_{1} G_{1}^{2}\right] \\ & =\frac{1}{16}\left[16 R^{2}-\left(A_{1} A_{2}^{2}+A_{1} A_{3}^{2}+A_{1} A_{4}^{2}\right)-\left(A_{2} A_{3}^{2}+A_{2} A_{4}^{2}+A_{3} A_{4}^{2}\right)\right]. \end{aligned}$$

Since $O G^{2} \geqslant 0$, it follows that $\sum_{1 \leqslant k<j \leqslant 4} A_{k} A_{j} \leqslant 16 R^{2}$.
(II) Clearly, $A_{1} G_{1} \geqslant h$, then $h_{1}^{2} \leqslant \frac{1}{9}\left[3\left(A_{1} A_{2}^{2}+A_{1} A_{3}^{2}+A_{1} A_{4}^{2}\right)-\left(A_{2} A_{3}^{2}+A_{2} A_{4}^{2}+A_{3} A_{4}^{2}\right)\right]$. Similarly, for $h_{2}, h_{3}, h_{4}$, we have similar inequalities.
Adding these four inequalities, we get $\sum_{1 \leqslant k<j \leqslant 4} A_{k} A_{j} \geqslant \frac{9}{4} \sum_{i=1}^{4} h_{i}^{2}$.
(III) By $V=\frac{1}{3} S_{i} h_{i}(i=1,2,3,4)$, we have $\sum_{i=1}^{4} \frac{1}{h_{i}}=\frac{1}{3 V} \sum_{i=1}^{4} S_{i}$.

Also, by $V=\frac{1}{3} \sum_{i=1}^{4} S_{i} \cdot r$, we have $\sum_{i=1}^{4} \frac{1}{h_{i}}=\frac{1}{r}$.
By $\sum_{i=1}^{4} \frac{1}{h_{i}} \geqslant 4\left(\frac{1}{h_{1} h_{2} h_{3} h_{4}}\right)^{\frac{1}{4}}$, we have $h_{1} h_{2} h_{3} h_{4} \geqslant(4 r)^{4}$.
Thus, $h_{1}^{2}+h_{2}^{2}+h_{3}^{2}+h_{4}^{2} \geqslant 4\left(h_{1} h_{2} h_{3} h_{4}\right)^{\frac{1}{2}} \geqslant 4\left[(4 r)^{4}\right]^{\frac{1}{2}}=(8 r)^{2}$.
(IV) Let the circumradius of tetrahedron $A^{\prime}{ }_{1} A^{\prime}{ }_{2} A^{\prime}{ }_{3} A^{\prime}{ }_{4}$ be $R^{\prime}$, then $R^{\prime 2} \geqslant \frac{1}{16} \sum_{1 \leqslant k<j \leqslant 4} A^{\prime}{ }_{k} A^{\prime}{ }_{j}$.

The circumradius of tetrahedron $A^{\prime}{ }_{1} A^{\prime}{ }_{2} A^{\prime}{ }_{3} A^{\prime}{ }_{4}$ is exactly the inradius of tetrahedron $A_{1} A_{2} A_{3} A_{4}$, so $R^{\prime}=r$. Thus, $R^{2} \geqslant \frac{1}{16} \sum_{1 \leqslant k<j \leqslant 4} A_{k} A_{j} \geqslant \frac{9}{64} \sum_{i=1}^{4} h_{i}^{2} \geqslant 9 r^{2}=9 R^{\prime 2} \geqslant \frac{9}{16} \sum_{1 \leqslant k<j \leqslant 4} A^{\prime}{ }_{k} A^{\prime}{ }_{j}$.
Therefore, $\sum_{1 \leqslant k<j \leqslant 4} A_{k} A_{j} \geqslant 9{ }_{1 \leqslant k<j \leqslant 4} A^{\prime}{ }_{k} A^{\prime}{ }_{j}$.