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 4 of 4: One of the two vertices must work

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In plain words

The proof is a two-way trap: either the first vertex passes immediately, or its failure supplies exactly the comparison needed to make the second vertex pass.

b+f>a \Longrightarrow\triangle(a,b,f),\qquad b+f\le a \Longrightarrow\ c+e>a \Longrightarrow\triangle(a,c,e)

Detailed analysis

There are only two cases. If $b+f>a$ , then the three edges $a,b,f$ satisfy all triangle inequalities and meet at one vertex. If $b+f\le a$ , Steps 2–3 give $c+e>a$ , and the three edges $a,c,e$ satisfy all triangle inequalities and meet at another vertex. In either case the tetrahedron contains a vertex whose three incident edge lengths are the sides of a triangle, proving the claim.