@Rouben Rostamian  Technically this is the Gateaux derivative (differential). Unlike the Frechet differential, it is not always linear.

@Kitonum A nice counterexample is the almost regular "tetrahedron" with sides [4,4,4,4,4,7].

@Kitonum I think I have already explained:
1. The orders are distinct. If you want the order xyzuvw, see the P.S. in my answer.
If you want exactly your order, change accordingly.
2. I have checked only the sorted sixtuples. To obtaind all the permutation, these permutations must be added, or, simply take permute instead of choose.

@Kitonum OK, but you have swiched the natural order xyzuvw to the unusual xyzwvu in your code. Just to fool me :-)
(and you were also fooled by my ordering).

@minhthien2016 I have considered only sorted sixtuples. To obtain all of them we must check the permutations of L in the for loop too. But there are too many!

P.S. Why did you detete the extra question in the meantime?

@Kitonum 

min(x+y+z,x+v+w,y+w+u,z+u+v) > max(x+u,y+v,z+w)

(x=DA, y=DB, z=DC, u=BC, v=CA, w=AB in the tetrahedron ABCD).

@Kitonum In your solution, the triangle inequalities are incorrect (you have typos).
[ 3, 5, 13, 7, 11, 17] is NOT a correct tetrahedron, (5,13,7)  cannot be the sides of a triangle.
(Note also that  we have different orderings and I have considered only sorted sixtuples) .

@Kitonum What is the order of the vertices? In my version, the lengths are [AB, AC, BC, DA, DB, DC].

@sursumCorda Thank you, I missed (or forgot?) the "Expression sequence" paragraph in the "Element-wise" help file.

@dharr 

for a=b=c=1 there is the valid solution x=y=z=k, for any k<>0.

@Alfred_F There is no chance to use GroupTheory. We would need RingTheory which Maple is missing.
But the idea was to show that Maple can be used to find the result and its proof, in case we don't know it.

@mmcdara Nice and much faster execution.
Note that my intention was to minimize: time(think) + time(code) + time(execute).

@Alfred_F The proof I know uses a Hamel basis. If we are given such a basis, the proof is "constructive", but for the basis itself the Axiom of Choice is needed.

@JAMET Actually the fixed point is the symmetric w.r.t. O  of the projection:

solve({x/a+y/b=1, a*x-b*y=0},{x,y});

        {x = a*b^2/(a^2 + b^2), y = b*a^2/(a^2 + b^2)}

@dharr Unit triangles are not enough.
Actually I forgot to mention that the 8 points are also not enough, but 10 are.

For example:  

Here monochromatic unit triangles do not exist.
If we remove the upper vertex to obtain a 9 point configuration, there are no monochromatic triangles!