- #1
James4
- 14
- 0
Hi
Given is a triangle on points x,y,z in the plane. This triangle has two points a and b on opposite sides (see Figure).
I would like to show that the following inequality has to hold:
\max {d(b,x), d(b,y), d(b,z)} +
\max {d(a,x), d(a,y), d(a,z)} - d(b,a)
> \min {d(x,y), d(x,z), d(y,z)}
where d(u,v) denotes the euclidean distance between u and v.
I actually expect the above statement to be true even if a and b are two arbitrary points outside of the triangle.
Does anybody have an idea how to approach this?
Given is a triangle on points x,y,z in the plane. This triangle has two points a and b on opposite sides (see Figure).
I would like to show that the following inequality has to hold:
\max {d(b,x), d(b,y), d(b,z)} +
\max {d(a,x), d(a,y), d(a,z)} - d(b,a)
> \min {d(x,y), d(x,z), d(y,z)}
where d(u,v) denotes the euclidean distance between u and v.
I actually expect the above statement to be true even if a and b are two arbitrary points outside of the triangle.
Does anybody have an idea how to approach this?
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