Let AB be the longest edge. All the other 4 points must then lie within the intersection of the circle with centre at A and radius AB, and the circle with centre B and radius AB. Let the point C be the next furthest point from A or B, WLOG let AC>BC. AB is the longest edge, so no point can be further than AB from point C. Also, BC is the shortest side in ABC, so no point can be within the intersection of the circles with radius BC and centres at B and C (or there would be a triangle whose longest edge is BC). This leaves an area such as the shaded area shown in the diagram. We then separate this region using the perpendicular bisectors of AB, BC and AC. At least two of D,E,F (say points D and E) must be on the side of the perpendicular bisector of AB closer to A, such that BD>AD and BE>AE. They must also be spaced far enough apart such that DE>EB or DE>DB. This is because if AD (or AE) were the longest side in ADE, it would be the shortest side in the triangle ADB (or AEB). Therefore we must have DE>BD or DE>BE. This leaves the only remaining possible spot for F being between B and D. Now consider triangle ACF. By assumption AC>AF, so the minimum side in this triangle can only be AF or CF. But AF is the longest side in triangle ADF, and CF is the longest side in BCF. So however the remaining points are arranged, there must be at least one triangle whose longest side is the shortest side in one of the other triangles.

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There is another solution using R(3,3)=6. Since the distances are all distinct we can colour tge edges in the following way: we look at an edge. If there is some triangle in which it has the shortest length then we colour it red, otherwise we colour it blue. Using now R(3,3)=6, we have some monochromatic triangle. Since its shortest edge must be red, all 3 edges are red. Looking now at the longest edge of tgis triangle we know it is red, so there is some triangle in which it has the shortest length q.e.d