Prove: d(x,z) <= d(x,y) + d(y,z) in R^2

[G-X]

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

0 = 0-vector

[MATH]d(x, y) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}[/MATH]

I was able to find and use [MATH](x_1z_2 - x_2z_1) \geq 0[/MATH]
To form:
[MATH] d(x,z) \leq d(x, 0) + d(z, 0) [/MATH]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 [MATH]d(x, y) + d(y, z) \leq d(x, 0) + d(y, 0) + d(y, 0) + d(z,0)[/MATH] and [MATH]d(x, z) \leq d(x, 0) + d(z, 0)[/MATH].

Any hints from here?

 

[Steven G]

d(x,z)≤d(x,y)+d(y,z) in R².

Shifting the problem to where x=0 should make things easier

 

[G-X]

I was able to figure out the solution thank you. Here are some more helpful hints. You can either think with regards to the triangle inequality of using a + b choosing a1,a2,b1,b2 or from d(x+y,0)≤d(x,0)+d(y,0) use symmetry which leads to the same solution. One quicker than the other.