Proof of an inequality in a metric space

Question

Problem:

Suppose that $d$ is a metric on a set $X$. Prove that the inequality $|d(x,y) - d(z,w)| \le d(x,z) + d(y,w) $ holds for all $w,x,y,z \in X$.

My attempt:

$ |d(x,y) - d(z,w)| \le d(x,z) + d(y,w)$

$d(x,y) - d(z,w) \le d(x,z) + d(y,w)$

$d(x,y) \le d(x,z) + d(z,w) + d(y,w) \le d(x,z) + d(y,w) $

Struck here.

Can anybody help me out.

Answer 1

Note that\begin{align}d(x,y)-d(z,w)&\leqslant d(x,z)+d(z,w)+d(w,y)-d(z,w)\\&=d(x,z)+d(w,y)\\&=d(x,z)+d(y,w).\end{align}Now, switching $x$ with $z$ and $y$ with $w$, you get that$$d(z,w)-d(x,y)\leqslant d(z,x)+d(w,y)=d(x,z)+d(y,w).$$Since the number $\bigl|d(x,y)-d(z,w)\bigr|$ is one of the numbers $\pm\bigl(d(x,y)-d(z,w)\bigr)$ and both of them are smaller than or equal to $d(x,z)+d(y,w)$, we have$$\bigl|d(x,y)-d(z,w)\bigr|\leqslant d(x,z)+d(y,w).$$