I did not do this inequality, but since you seem knowledgable this should greatly help you finish it.Originally Posted by malaygoel
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
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, the "Fermat Point" form lines of 120 degrees.