How does this inequality hold? $|d(x,x_0)-d(y,x_0)|\le d(x,y)$

Question

I came across this during some reading: $|d(x,x_0)-d(y,x_0)|\le d(x,y)$. I can't seem to figure out why it holds. Here $d$ is a metric.

Answer 1

Considering the triangle inequality we know that:

$$ d(x,x_0) \leq d(x,y) + d(y,x_0) $$

Can you use this fact to achieve that $ |d(x,x_0) - d(y,x_0)| \leq |d(x,y)| $ ? Once you know this, recall that $d$ is a non-negative function...

Answer 2

Combine the following two applications of the triangle inequality: \begin{align} d(x,x_0) &\le d(x,y) + d(y, x_0) \\ d(y, x_0) &\le d(x,y) + d(x, x_0) \end{align}

Answer 3

HINT: Use the triangle inequality to show that $d(x,y)\ge d(x,x_0)-d(y,x_0)$ and $d(x,y)\ge d(y,x_0)-d(x,x_0)$.