Can You Prove the Inequality in Triangles with Arbitrary Points?

[James4]

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?

 

[chiro]

Hey James4.

Have you considered using the triangle inequalities?

http://en.wikipedia.org/wiki/Triangle_inequality

 

[James4]

Hi chiro

Thanks, for your answer.
Yes I have considered this, but I don't think that they immediately apply to the problem.

 

[chiro]

Can you consider using the inequality to get the maximum of each distance function? (Think about the triangle with vertices x,y,z and apply the inequality to the arbitrary point b).

 

[James4]

Hi

I am not sure if I understand what you mean by "get the maximum of each distance function".
Do you mean distinguishing the cases when xy, xz or yz are is the largest side?

 

[chiro]

You have a max(d(x,b),d(y,b),d(z,b)) and I was referring to that specific function itself.

 

[James4]

thanks, but how do you know which side is the largest? For example in the triangle y,b,x, any side can be largest, depending on the angles in the original triangle.
So I don't see which bounds you would obtain.

Btw: Do you have a solution in mind and you want to guide me there or are you also thinking about how to solve this problem? Because I don't think it is trivial.

 

[chiro]

I'm just bouncing ideas off you to help you solve your own problem: I didn't intend to solve the whole thing completely for you.