Prove the triangle inequality in R^2 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 $\sqrt{(x_1-z_1)^2 + (x_2-z_2)^2}$.
 A: 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.
A: Show this first when one of the points is the origin. Then reduce all other cases to this case by using a suitable translation.
