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 2 of 4: Assume one candidate fails; the opposite face gives e > b

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In plain words

The failed inequality $b+f\le a$ is not wasted: when compared with the strict face inequality $e+f>a$ , it immediately tells us that $e$ must be longer than $b$ .

b+f\le a \;\Longrightarrow\; e+f>a \;\Longrightarrow\; e>b

Detailed analysis

If $b+f>a$ , then the edges $a,b,f$ already have their largest edge smaller than the sum of the other two, and the other two triangle inequalities follow from $a$ being the maximum: $a+b>f$ and $a+f>b$ . So suppose instead that $b+f\le a$ . In the face whose edges are $a,e,f$ , the triangle inequality gives $e+f>a$ . Combining $b+f\le a<e+f$ and cancelling $f$ yields $e>b$ .