6. The minimum value of the sum of the distances from a point in space to the vertices of a regular tetrahedron with edge length 1 is
A. $\sqrt{6}$
B. $2 \sqrt{3}$
C. 4
I). $2 \sqrt{2}$
6. Let the edge length of the regular tetrahedron $ABCD$ be $1$, and let $E$ be the midpoint of $BC$. Then the plane $ADE$ is the perpendicular bisector plane of the line segment $BC$.
Let the projection of $P$ on the plane $ADE$ be $H$, then $PA \geqslant HA, PD \geqslant HD$,
and $PH / / BC, HB=HC$, so $PB + PC \geqslant HB + HC$.
Therefore, $PA + PB + PC + PD \geqslant HA + HB + HC + HD$.
Thus, the point $P$ that minimizes $PA + PB + PC + PD$ lies on the perpendicular bisector plane of the line segment $BC$. Similarly, the point $P$ that minimizes the sum also lies on the perpendicular bisector planes of the other edges,
hence this point can only be the center $O$ of the regular tetrahedron $ABCD$. It is easy to see that $AO = \frac{\sqrt{6}}{4}$, so the minimum value of $PA + PB + PC + PD$ is $\sqrt{6}$.