matematikros wrote:

This explanation does not seem right. because, subject to all of this tetrahedron $ABCD$ is regular tetrahedron. With your solution, just a special case of the tetrahedron $ABCD$, not true for the general:$MA+MB+MC+MD \ge 4R$.

matematikros wrote:

Why does this hold??

because quadratic mean greater than or equal to algebric mean

But I wonder why he make equivalent of AM with QM

I'm pretty sure the above solution is incorrect, since AM<=QM, not the other way around. Let $ABCD$ be the tetrahedron and $P$ be any point in space. We want to find the optimal point $P$. We let $P$ vary on the plane passing through it parallel to $ABC$. Then obviously, the distance $PD$ is minimal when $P$ lies on line $DO$, where $O$ is the centroid of the tetrahedron. We claim that $PA+PB+PC$ is then minimal as well, which will finish the problem (since then $P$ analogously lies on $BO$, so $P=O$). To show this, we now fix $PC$, so that $P$ varies on a fixed circle, the center of which is equidistant from $A,B$. Now consider the set of points $M$ for which $AM+BM$ is fixed. This is an ellipsoid and thus, a convex set. We now consider growing ellipsoids and the minimum $AP+BP$ for $P$ on the circle is obviously achieved for the point where the ellipsoid is first tangent to the circle. Assume that this point does not lie symmetrically with the plane $\mathcal P_{CDO}$. Then we could reflect the minimum across this plane. By symmetry, this point has the same sum of distances. By convexity, their midpoint lies inside the ellipsoid and thus, it has a smaller sum of distances. However, the point where $\mathcal P_{CDO}$ intersects the circle (the intersection point closer to $A,B$), is obviously closer to $A,B$ than this midpoint. Hence, we conclude $O$ is indeed the desired point.