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You are potentially able to prove $d(x,z) \leq d(x,y) + d(y,z)$ in $\textbf{R}^2$ using the relation that $d(x + y, 0) \leq d(x, 0) + d(y, 0)$ where:

$0 = 0$-vector

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

I was able to find and use $(x_1z_2 - x_2z_1) \geq 0$

To form $d(x,z) \leq d(x, 0) + d(z, 0)$ then by symmetry \begin{align*} \begin{cases} d(x,y) \leq d(x, 0) + d(y, 0)\\\\ d(y,z) \leq d(y, 0) + d(z, 0) \end{cases} \end{align*}

You get $d(x, y) + d(y, z) \leq d(x, 0) + d(y, 0) + d(y, 0) + d(z,0)$ and $d(x, z) \leq d(x, 0) + d(z, 0)$.

Any hints from here?

user806208
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  • Cauchy-Schwarz inequality has yet to be defined. I think your suppose to be able to use: $d(x + y, 0) <= d(x, 0) + d(y, 0) in some manner. – user806208 Jul 05 '20 at 16:45

1 Answers1

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$d(x,z)=d(x-z,0)=d(x-y+y-z,0)\leq d(x-y,0)+d(y-z,0)=d(x,y)+d(y,z)$.

NeitherNor
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  • This post got me working towards a correct solution.

    You can either think with regards to the triangle inequality of using a + b or from $d(x + y, 0) \leq d(x, 0) + d(y, 0)$ use symmetry which leads to the same solution. One quicker than the other.

    A question that you will recognize as having similar properties is: https://math.stackexchange.com/questions/2122350/proving-the-triangle-inequality-for-the-euclidean-distance-in-the-plane

     – user806208 Jul 07 '20 at 20:53