For regular tetrahedron $ABCD$ with center $O$, and $\overrightarrow{NO}=-3\overrightarrow{MO}$, is $NA+NB+NC+ND\geq MA+MB+MC+MD$?

Solution 1:

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.

Solution 2:

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.