7.1. Given a rectangular parallelepiped $2 \times 3 \times 2 \sqrt{3}$. What is the smallest value that the sum of the distances from an arbitrary point in space to all eight of its vertices can take?

![](https://cdn.mathpix.com/cropped/2024_05_06_b6fc33c08418efc8d3bcg-2.jpg?height=394&width=348&top_left_y=1982&top_left_x=243)
Answer: 20

Solution. Let $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be the given parallelepiped. The sum of the distances from point $X$ to two opposite vertices $A$ and $C^{\prime}$ is not less than the length of $A C^{\prime}$, i.e., the length $d$ of the major diagonal. Summing up similar inequalities, we have $X A+X B+X C+X D+X A^{\prime}+X B^{\prime}+X C^{\prime}+X D^{\prime}=\left(X A+X C^{\prime}\right)+\left(X B+X D^{\prime}\right)+$ $+\left(X C+X A^{\prime}\right)+\left(X D+X B^{\prime}\right) \geqslant A C^{\prime}+B D^{\prime}+C A^{\prime}+D B^{\prime}=4 d$. Equality is achieved when $X$ is the center of the parallelepiped.

It remains to calculate $d=\sqrt{A B^{2}+A D^{2}+A A^{\prime 2}}=5$, and then the answer is $4 d=20$.