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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 $\sqrt{(x_1-z_1)^2 + (x_2-z_2)^2}$.

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

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  • $\begingroup$ 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
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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.

ANSWER:

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.

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