I need to prove that $d(y,z) + d(x,y) \geq d(x,z)$.

With $d(x,y) = \sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}$.

I'm struggling to figure out how to work with the square roots especially when they're over an addition problem. Any help is appreciated.

Edit: to make this simpler. I think all I need to do is figure out how to add $\sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}$ + $\sqrt{(y_1-z_1)^2 + (y_2-z_2)^2}$. Then I can do the a + b proof method. But I just can't figure out how to add those equations so it ends up equalling $\sqrt{(x_1-z_1)^2 + (x_2-z_2)^2}$.

  • $\begingroup$ Euclid gave a very nice proof in his Elements: en.wikipedia.org/wiki/Triangle_inequality. $\endgroup$ – Chris Custer Jan 18 '19 at 2:20
  • $\begingroup$ In one dimension, the formula states that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides. $\endgroup$ – herb steinberg Jan 18 '19 at 2:21
  • $\begingroup$ @herbsteinberg I need to do it in two dimensions though. That's my problem. With one dimension I was able to easily see that (x - y) + (y-z) = (x-z) but having to use the distance formula for two dimensions is making it much more difficult to show that as true. $\endgroup$ – user580909 Jan 18 '19 at 2:28
  • $\begingroup$ @user580909 I shouldn't have said one dimension. I simply meant the triangle inequality, which is best proven geometrically not algebraically. $\endgroup$ – herb steinberg Jan 18 '19 at 2:37
  • $\begingroup$ @herbsteinberg I don't think I made this clear but I don't have a problem proving the triangle inequality. My problem is that my proof rests on d(x,y) + d(y,z) = d(x,z) in the second dimension. That's what I need to prove so that I can then use a proof involving abs(a) + abs(b) = abs(a+b) $\endgroup$ – user580909 Jan 18 '19 at 2:41

Show this first when one of the points is the origin. Then reduce all other cases to this case by using a suitable translation.

  • $\begingroup$ Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax. $\endgroup$ – dantopa Jan 18 '19 at 3:00

The OP is attempting to define a metric on the set of points in the Cartesian plane, $\mathbb R \times \mathbb R$. It can be safely assumed that before arriving at this point, they have been introduced to the idea of calculating the length of vectors.

In any case, I doubt there is direct algebraic technique showing

$\tag 1 \sqrt{(x_1-z_1)^2 + (x_2-z_2)^2} \le \sqrt{(x_1-y_1)^2 + (x_2-y_2)^2} + \sqrt{(y_1-z_1)^2 + (y_2-z_2)^2} $

without taking a journey upon which you discover the Cauchy–Schwarz inequality.

So when the OP states

I think all I need to do is figure out...

they are bound for disappointment.

But what an opportunity to reflect on this amazing mathematical material. The inequality $\text{(1)}$ is true for all $x_1, x_2,y_1,y_2,z_1,z_2 \in \mathbb R$, and it is still true if we 'only live on the line'. But so much abstract modern mathematical thought must be expended to bring it all (i.e. Cartesian Coordinate Space = Euclidean Space) to life.


Using the C-S inequality,

$\tag 2 (u_1 v_1 + u_2 v_2)^2 \le (u_{1}^{2}+ u_{2}^{2})\,(v_{1}^{2}+ v_{2}^{2})$

among other arguments, is the way to go if you want to show that $d(u,v)$ satisfies the triangle inequality.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.