Triangle Inequality (Sums of Sides)

[SpringPhysics]

Homework Statement

Prove that any side of a triangle is less than or equal to the sum of the other two sides of the triangle (using components).

Homework Equations

root [(x3-x1)^2 + (y3-y1)^2]
<= root [(x2-x1)^2 + (y2-y1)^2] + root [(x3-x2)^2 + (y3-y2)^2]

The Attempt at a Solution

I attempted to solve the inequality portion by considering the slopes of the sides of the triangle (using components), where

[(x2-x1)/(y2-y1)]^2 + [(x2-x1)/(y2-y1)]^2 + [(x2-x1)/(y2-y1)]^2 > 0
I expanded everything but that got really messy. We haven't learned to really prove anything so far (but the textbook only mentions the distance formula), so I don't really know where to start with the problem. Our professor showed us a method that involved squaring differences of numbers in order to prove something (and taking the discriminant), but it doesn't seem to be viable here.

 

[slider142]

The slopes have very little to do with the triangle inequality, which is a fundamental in how mathematicians generalize the property of measuring "length", using the assumption that "length" is always positive.
Use the distance formula on the components. Without loss of generality, you can place one vertex of the triangle at the origin. Then draw a line to the arbitrary point (x0, y0). Another arbitrary point (x1, y1) is needed, but the completing line segment is now fully defined. Noting that the distance formula is always positive and therefore the sum of distances is positive as well, see if this gives you the inequality you seek. Afterwards, note that the placement of the origin is arbitrary and add the (x3, y3) point back into the formula. This is done to save the forests.

 

[SpringPhysics]

Even so, the three points may be collinear (therefore the slopes of the segments are the same) and the sum of the two smaller segments would equal the third segment. That's why I thought that finding slopes (and suggesting that the slopes are different) would be the method for proving the inequality portion of the inequality (sorry if it sounded like I didn't know the equality, which I do).

 

[slider142]

Even so, the three points may be collinear (therefore the slopes of the segments are the same) and the sum of the two smaller segments would equal the third segment. That's why I thought that finding slopes (and suggesting that the slopes are different) would be the method for proving the inequality portion of the inequality (sorry if it sounded like I didn't know the equality, which I do).

that's why there is an = sign in the inequality as well. ;) Did you solve this one yet?

 

[SpringPhysics]

I solved the equality but not the "<" part of it. I would think we would have to compare slopes in order to prove it because otherwise all three points could lie on the same line.

 

[slider142]

I solved the equality but not the "<" part of it. I would think we would have to compare slopes in order to prove it because otherwise all three points could lie on the same line.

Again, the slopes have very little to do with the inequality. The inequality comes right from writing the sum of the three lengths of the legs of a triangle and then noting that this sum is greater than or equal to 0. Basic inequality algebra then gives you the triangle inequality. The only ingredient is the distance formula.

 

[SpringPhysics]

Isn't that a little simple for a proof though?

(Thanks for putting up with all of this.)

 

[slider142]

Isn't that a little simple for a proof though?

(Thanks for putting up with all of this.)

I'm not sure what you mean. Simple proofs are to be encouraged; they show how fundamental a concept is.

 

[SpringPhysics]

It's just that the solution seems to be rephrasing the question. And when you bring a term over from the sum, wouldn't that term be negative? Would you then say that negative distance is invalid and so QED?

 

[slider142]

It's just that the solution seems to be rephrasing the question. And when you bring a term over from the sum, wouldn't that term be negative? Would you then say that negative distance is invalid and so QED?

No, you then square both sides. It's not *that* simple. :D

 

[SpringPhysics]

I thought about that too. Okay then, thanks for your help!