当P位于ABC平面之外时，Pompeiu定理仍成立？

hbghlyj

设P为等边三角形ABC平面上的一点，托勒密不等式推出线段AP、BP和CP的长度为一个三角形的边。
当P位于ABC的外接圆上时，该三角形退化。

在$\mathbb R^3$中，当P位于ABC平面之外时，Pompeiu定理仍成立，如何证明？

hbghlyj

doi.org/10.2307%2F3616345
THEOREM. The product of any couple of opposite edges of a tetrahedron is less than the sum of the products of the two others.
Without loss of generality, that is $e f<a c+b d$ (see Fig. 1), in the tetrahedron $A B C D$, where $A B=a, B C=b, C D=c, D A=d, A C=e$ and $B D=f$.

Rotate $\triangle C B D$ around the axis $B D$ (see Fig. 3) so that $\triangle B D C_1$ is congruent to $\triangle B D C$ and $A, B, D, C_1$ are planar. Let $A C_1=e_1$ and let $A C_1$ and $B D$ intersect at $O$. Then clearly
\[
e \leqslant A O+O C=A O+O C_1=e_1
\]
with equality if and only if $C=C_1$. Hence, for a proper tetrahedron, $e<e_1$ and (from the above proposition)
\[
e f<e_1 f \leqslant a c+b d
\]