MathLabs

Problem 4

Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which are the sides of a triangle.
Step 1 of 4: Label the six edges and choose a longest one
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In plain words

The longest edge is the only possible obstruction: if two edges at a vertex failed to add up past it, that vertex would not work. The tetrahedron's other faces constrain the remaining edges so strongly that the obstruction can be transferred to a different vertex and disappears there.

{a,b,c}=edges of one face,{d,e,f}=opposite edges,a=max⁡{a,b,c,d,e,f}\{a,b,c\}=\text{edges of one face},\qquad \{d,e,f\}=\text{opposite edges},\qquad a=\max\{a,b,c,d,e,f\}
A tetrahedron with a small separation between faces, illustrating the four vertices and the three edges meeting at each vertex
A three-dimensional tetrahedron, shown with its faces slightly separated so that its four vertices and the three incident edges at each vertex are visible.
Detailed analysis

Take one triangular face and call its edge lengths a,b,ca,b,c. Call the three edges not adjacent to a,b,ca,b,c, respectively, d,e,fd,e,f. Relabel the tetrahedron if necessary so that aa is the maximum of all six edge lengths. The goal is to prove that either the three edges a,b,fa,b,f at one vertex or the three edges a,c,ea,c,e at another vertex satisfy all three strict triangle inequalities.