# Inequality in Triangles

1. Oct 25, 2012

### 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?

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Last edited: Oct 25, 2012
2. Oct 26, 2012

### chiro

3. Oct 26, 2012

### James4

Hi chiro

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

Last edited: Oct 26, 2012
4. Oct 27, 2012

### 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).

5. Oct 27, 2012

### 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?

6. Oct 27, 2012

### chiro

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

7. Oct 27, 2012

### 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.

8. Oct 27, 2012

### 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.