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 3 of 4: The other face forces c + e > a

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In plain words

One strict inequality from each of two faces combines in a monotone way: replacing $b$ by the larger edge $e$ can only increase the sum paired with $c$ , turning the failed candidate into a successful one.

b+c>a,\qquad e>b \;\Longrightarrow\; c+e>b+c>a

Detailed analysis

The original face with edges $a,b,c$ is a genuine triangle, so $b+c>a$ . From Step 2 we have $e>b$ . Therefore $c+e>c+b>a$ . Thus the three edges $a,c,e$ have the longest one $a$ smaller than the sum of the other two; and because $a$ is globally longest, the other two inequalities $a+c>e$ and $a+e>c$ are automatic.