I did not do this inequality, but since you seem knowledgable this should greatly help you finish it.Originally Posted bymalaygoel

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That point is called the "Fermat Point".

Consider triangle ABC and let P be inside this triangle.

We want that AP+BP+CP to be minimized.

Rotate 60 degrees triangle APB around point B to outside.

You now have a new triangle A'P'B

Triangle P'BP is equilateral cuz, P'B=PB with <P'BP=60

Thus AP+BP+CP=A'P'+P'P+PC

But A'P'+P'P+PC is the path from A' to C

It is minimal when A'P'+P'P+PC is a sraight line.

Thus, BPC=180-P'PB=180-60=120

Thus, the "Fermat Point" form lines of 120 degrees.