For regular tetrahedron $ABCD$ with center $O$, and $\overrightarrow{NO}=-3\overrightarrow{MO}$, is $NA+NB+NC+ND\geq MA+MB+MC+MD$? 
Let $ABCD$  be a regular tetrahedron with center $O.$ Consider two points $M,N,$ such that $\overrightarrow{NO}=-3\overrightarrow{MO}.$ Prove or disprove that 
  $$NA+NB+NC+ND\geq MA+MB+MC+MD$$

I tried to use CS in the Euclidean space $E_3$, but it does not help, because the minoration is too wide.
 A: This is just the $2$ dimensional variant of the same problem. 

Let $ABC$ be a equilateral triangle with center $O.$ Consider two points $M,N,$ such that $\overrightarrow{NO}=-2\overrightarrow{MO}.$ Prove that: 
  $$NA+NB+NC\geq MA+MB+MC.$$

Let $A',B',C'$ be the midpoints of the segments $BC, CA$ and $AB$ respectively. Notice that the map $$M\mapsto N$$ is actually a homothety with respect to $O$ and factor $-2$. 

So this homothety takes $X'\mapsto X$. We thus have $$NX = 2MX'$$
for each $X\in \{A,B,C\}$ and we have to prove $$2(MA'+MB'+MC')\geq MA+MB+MC.$$ 
Now observe that in a quadrilateral $AB'NC'$, by Ptolemy’s inequality, we have $$AB'\cdot MC'+AC'\cdot MB'\geq B'C'\cdot MA.$$ 

Notice that $AB'= AC' = B'C'$, so we have $$MC'+MB'\geq  MA.$$ And similarly we have $$MC'+MA'\geq  MB$$ and $$MA'+MB'\geq  MC.$$
Adding all three equations we get what we want to prove.
A: In Aqua's proof, we have to show that $MB' + MC' \geq MA$. I want to introduce another way :
In equilateral triangle $\Delta ABC$, we have mid points $A'$ in
$[BC]$ and we have $B',\ C'$ similarly.
When $O$ is in $\Delta A'B'C'$, then we have $O'$ s.t. $\Delta
O'B'C'$ is congruent to $\Delta OA'C'$. 
Hence $$ |A-O' | + |O'-O|  = |B'-O| + |O-C'| \geq |A-O| $$
This completes the proof.
A: Considering the distance squared, the problem becomes trivial. I leave this as a curiosity, not as an answer.
Given a regular tetrahedron centered at the origin as
$$
\cal{T}=r\left(
\begin{array}{ccc}
 \frac{2 \sqrt{2}}{3} & 0 & -\frac{1}{3} \\
 -\frac{\sqrt{2}}{3} & \sqrt{\frac{2}{3}} & -\frac{1}{3} \\
 -\frac{\sqrt{2}}{3} & -\sqrt{\frac{2}{3}} & -\frac{1}{3} \\
 0 & 0 & 1 \\
\end{array}
\right)
$$
a point $M = (m_x,m_y,m_z)$ and a rotation matrix $R(\alpha,\beta,\gamma)$ we have
$$
\sum_{k=1}^4(M-R\cdot {\cal{T}}(k))^{\dagger}\cdot (M-R\cdot {\cal{T}}(k)) =4||M||^2-2\sum_{k=1}^4M^{\dagger}\cdot R\cdot{\cal{T}}(k)+ \sum_{k=1}^4 ||{\cal{T}}(k)||^2
$$
but for the regular tetrahedron
$$
\sum_{k=1}^4M^{\dagger}\cdot R\cdot{\cal{T}}(k)=0\\
\sum_{k=1}^4 ||{\cal{T}(k)}||^2 = 4r^2
$$
then
$$
\sum_{k=1}^4(M-R\cdot {\cal{T}}(k))^{\dagger}\cdot (M-R\cdot {\cal{T}}(k)) = 4\left(||M||^2+r^2\right)
$$
independent of $\{\alpha,\beta,\gamma\}$
NOTE
Considering now the sum of distances as required, we have
$$
||\vec{MO}(\alpha,\beta,\gamma)||=\sum_{k=1}^4||M-R\cdot {\cal{T}}(k))|| =\sum_{k=1}^4\sqrt{||M||^2-2M^{\dagger}\cdot R(\alpha,\beta,\gamma)\cdot{\cal{T}}(k)+ r^2}
$$
and now the question
Is
$$
\sum_{k=1}^4\sqrt{9||M||^2+6M^{\dagger}\cdot R(\alpha,\beta,\gamma)\cdot{\cal{T}}(k)+ r^2}\ge \sum_{k=1}^4\sqrt{||M||^2-2M^{\dagger}\cdot R(\alpha,\beta,\gamma)\cdot{\cal{T}}(k)+ r^2}
$$
for all $\{\alpha, \beta, \gamma\}$ ?
Follows a MATHEMATICA script which minimizes
$$
\Delta(\alpha,\beta,\gamma)=\sum_{k=1}^4\sqrt{9||M||^2+6M^{\dagger}\cdot R(\alpha,\beta,\gamma)\cdot{\cal{T}}(k)+ r^2}-\sum_{k=1}^4\sqrt{||M||^2-2M^{\dagger}\cdot R(\alpha,\beta,\gamma)\cdot{\cal{T}}(k)+ r^2}
$$
for values of $\{\alpha,\beta,\gamma\},\ \rho = ||\vec{MO}||/r$ characterizing numerically the assumption
$$NA+NB+NC+ND\geq MA+MB+MC+MD$$
M = rho {1, 0, 0};
R = EulerMatrix[{alpha, beta, gamma}];
tetraR = Table[R.tetra[[k]], {k, 1, 4}];
sd = Sum[Sqrt[(-3 M - tetraR[[k]]).(-3 M - tetraR[[k]])], {k, 1, 4}] - Sum[Sqrt[(M - tetraR[[k]]).(M - tetraR[[k]])], {k, 1, 4}];
list = Table[{rhok, First[NMinimize[(sd /. {rho -> rhok}), {alpha, beta, gamma}]]}, {rhok, 0, 0.5, 0.005}];
ListLinePlot[list, PlotStyle -> {Blue, Thick}]

Follows the plot of $\Delta(\alpha^*,\beta^*,\gamma^*)\times\rho$

Follows the relative positions for given  $M$, $N$, and $\{\alpha^*,\beta^*,\gamma^*\}$

A: Another solution for two-dimensional problem. We will change notations to make them more logical.

Problem. Let $ABC$ be an equailateral triangle with center $O$. Consider two points $P$ and $K$, such that $\overrightarrow{OP}=-2\overrightarrow{OK}$. Prove that $PA+PB+PC\geq KA+KB+KC$.

Solution. Construct an equilateral triangle $PQR$ centered at $O$. Then, $K$ is the midpoint of $QR$ (due to $\overrightarrow{OP}=-2\overrightarrow{OK}$). Denote midpoints of $PQ$ and $RP$ as $M$ and $L$, respectively. Now our construction is "symmetric" (i. e. invariant under rotations by $\frac{2\pi k}{3}$, $k=0,1,2$). Hence, we have the following equalities
$$
PB=QC=RA,
\\
PC=QA=RB,
$$
and
$$
KB=LC=MA,
\\
KC=LA=MB.
$$
Thus, our inequality can be rewritten as 
$$
PA+QA+RA\geq KA+LA+MA.
$$
Finally, note that $K$ is the midpoint of $QR$, so $PA+QA\geq 2KA$. Adding similar inequalities gives the desired inequality.
Further thoughts. I tried to generalize this approach for the original problem but didn't succeed. The problem is in constructing regular tetrahedron $PQRS$. (It seems to me that it's connected to existence of normal subgroup of order 4 in $A_4$) 

Remark. The solution was taken from the post on AoPS (it was the problem on the China West MO 2014, by the way). There are two solutions one of which is similar to Aqua's one.
https://artofproblemsolving.com/community/c6h602780p3578945
