Prove the equation $\vert d(x,z)-d(y,t)\vert\leq d(x,y)+d(z,t)$

Question

I know how it verified the following equation: $$\vert d(x,y)-d(x,z)\vert\leq d(y,z)$$ where $x,y,z$ is arbitrary points of metric space $(X, d)$

But I didn't now how to prove the follow equation: $$\vert d(x,z)-d(y,t)\vert\leq d(x,y)+d(z,t)$$ where $x,y,z, t$ is arbitrary points of metric space $(X, d).$

I hope you will help me. Thank you very much for your help.

Answer 1

You can see it directly with:

$$d(x,z) \leq d(x,y)+d(y,t) + d(t,z)\implies\\d(x,z)-d(y,t)\leq d(x,y)+d(t,z)$$

and $$d(y,t)\leq d(y,x) + d(x,z)+d(z,t) = d(x,y)+d(x,z)+d(t,z)\implies \\d(y,t)-d(x,z)\leq d(x,y)+d(t,z)$$

Answer 2

So you know $$|d(x,z)-d(z,y)|\leq d(x,y)\text{ and }|d(z,y)-d(y,t)|\leq d(z,t).$$

Can you continue from these two inequalities?

Answer 3

\begin{align} &|d(x,z) - d(y,t)|\\ = &|d(x,z) - d(z,y) + d(z,y) - d(y,t)| \\ \leq & |d(x,z) - d(z,y)| + |d(z,y) - d(y,t)|\\ \leq & d(x,y) + d(z,t) \\ \end{align}