Tetra Inequalities

Well done to Gabriel from London Academy of Excellence Tottenham and Ruth from Manchester High School for Girls who solved this problem correctly. Gabriel sent in a long exploration and Ruth sent in a short proof. Here, Gabriel's exploration is shown first and Ruth's proof is shown below.

First, Gabriel explored the properties of triangles, and what it means if three lengths 'could be the sides of a triangle' (click to enlarge):

Next, Gabriel applied these ideas to a tetrahedron, and assumed that none of the vertices are joined to three edges which could be the sides of a triangle (click to enlarge):

Gabriel then made another assumption, which allowed the proof to be completed:

It was fair for Gabriel to assume that $a$ was the longest side, because at that point there was nothing to differentiate between the sides - any of them could have been called $a,$ so we can choose that $a$ is the longest side.

Once $a$ is the longest side, there is now a diffrerence between the other sides: they either share an edge with $a$ (ie $b,c,d$ or $e$) or they do not share an edge with $a$ (ie $f$). Gabriel has proved that if $f$ is the second-longest side, then the tetrahedron is impossible. What if $b,c,d$ or $e$ is the second longest side?

Here is Ruth's proof that in every tetrahedron there is a vertex where the three edges meeting at that vertex have lengths which could be the lengths of the sides of a triangle:

The tetrahedron's vertexes are $A, B, C$ and $D$ and the longest side is $AB.$
If you assume that there is not a vertex where the three sides meeting at it could be the sides of a triangle, 

$AB + \text{either of the other sides at }A\text{ or }B \gt \text{the third side at that vertex}$
$\therefore AC + AD \lt AB$ and $BC + BD \lt AB,$ otherwise the sides meeting at $A$ or $B$ could be the sides of a triangle.
$\therefore AC + AD + BC + BD \lt 2AB\hspace{6mm}(1)$

$ABC$ and $ABD$ are both triangles
$\therefore AC + BC \gt AB$ and $AD + BD \gt AB$
$\therefore AC + AD + BC + BD \gt 2AB\hspace{6mm}(2)$

$(1)$ and $(2)$ are contradictions
$\therefore$ the initial assumption is wrong
$\therefore$ there is at least one vertex where the three sides meeting at it could be the sides of a triangle.

Can you represent the triangle inequality using a diagram? Can you explain Ruth's proof using a diagram?