
> 
> 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.  

Note: I also posted this [on the Mathematics Stack Exchange](https://math.stackexchange.com/questions/3485563/for-regular-tetrahedron-abcd-with-center-o-and-overrightarrowno-3-over), but not much progress has been made on this question. This is why I thought that posting here too would be all right (this problem is open in the sense that its proposer doesn't have a proof, so I guess it is fit the for this forum).  

EDIT: The bounty expired, so this may be reopened.



For regular tetrahedron $ABCD$ with center $O$, and $\overrightarrow{NO}=-3\overrightarrow{MO}$, is $NA+NB+NC+ND\geq MA+MB+MC+MD$?
Following suggestions on Stack Exchange, we use the homothety with respect to $O$
 and factor $-3$.
If $X$ is a point, then $X'$ denotes the point for which $\overrightarrow{XO}= -3 \overrightarrow{X'O}$.
So $M=N'$ and $A'$ is the midpoint of the face opposite to $A$.
One has $XY=3X'Y'$ and the desired inequality becomes
$3MA'+3MB'+3MC'+3MD'\geq MA+MB+MC+MD$.


It thus suffices to prove inequalities of the type $MB'+MC'+MD'\geq MA$.
Thus put $f(X)=XB'+XC'+XD'-XA$. We have to show that $f(M)\geq0$.


Let us fix our tetrahedron as follows.


$A=\left(0,0,\sqrt{\frac{2}{3}}-\frac{1}{2
 \sqrt{6}}\right)$,


$B= \left(-\frac{1}{2 \sqrt{3}},-\frac{1}{2},-\frac{1}{2
 \sqrt{6}}\right)$,


$C=\left(-\frac{1}{2 \sqrt{3}},\frac{1}{2},-\frac{1}{2
 \sqrt{6}}\right)$,


$D=\left(-\frac{1}{2 \sqrt{3}},\frac{1}{2},-\frac{1}{2
 \sqrt{6}}\right)$.


Put $f\_1(X)=XB'+XC'$, $f\_2(X)=XD'-XA$, so that $f(X)=f\_1(X)+f\_2(X)$.


Notice that 
$f(O)=0$ and notice that $f(M)$ is positive if $M$ is far away from $O$.


Now take $M$ so that $f(M)$ is an absolute minimum.
Note that $f$ is positive at $A$, $B'$, $C'$, $D'$, so that $M$ is none of those points.
We have $f\_2(M)<0$ and the gradients of $f\_1$, $f\_2$ cancel each other at $M$.
That means that the level sets through $M$ of $f\_1$ and $f\_2$ touch at $M$. (There is no point where both gradients vanish.)


The level set through $M$ of $f\_1$ is an ellipsoid, with $f\_1$ smaller inside, and the level set through $M$ of $f\_2$ is the lower sheet 
of a hyperboloid of two sheets, with $f\_2$ more negative inside the sheet.
As each level set is the curved boundary of a convex region, there is no other
point where the two level sets touch.


The level sets are symmetric with respect to the plane $ \left\{X\mid XB'=XC'
 \right\}$, so $M$ must lie on that plane. We have shown that $MB'=MC'$.
 Similarly $MC'=MD'$ and $M$ must be of the form $(0,0,x)$, so that
 $f(M)=\frac{\sqrt{72 x^2-4 \sqrt{6} x+3}-\sqrt{8 x^2-4
 \sqrt{6} x+3}}{2 \sqrt{2}}\geq0$.


