How to prove that $|d(x,y)-d(z,w)|\leq |d(x,z)+d(y,w)|$? How to use the triangle inequality to prove that $|d(x,y)-d(z,w)|\leq |d(x,z)+d(y,w)|$?
 A: Combine the following two applications of the triangle inequality.
\begin{align}
d(x,y)
&\le d(x,z) + d(z,w) + d(y,w)\\
d(z,w)
&\le d(z,x) + d(x,y) + d(y,w)
\end{align}

Explicitly, rearranging each inequality above gives
\begin{align}
d(x,y) - d(z,w) &\le d(x,z) + d(y,w)\\
d(z,w) - d(x,y) &\le d(x,z) + d(y,w).
\end{align}
This implies
$$|d(z,w) - d(x,y)| \le d(x,z) + d(y,w).$$
Finally, the right-hand side is always positive, so you can include absolute value signs if you want.
A: Use
$$d(x,y)\leq d(x,z)+d(y,w)+d(z,w)\implies d(x,y)-d(z,w)\leq d(x,z)+d(y,w)$$
and
$$d(z,w)\leq d(z,x)+d(x,y)+d(y,w)\implies d(z,w)-d(x,y)\leq d(z,x)+d(y,w)$$
to obtain the desired result.
A: It may help to draw a picture.
Let $d(x,y)=a, d(w,z)=b, d(x,z)=c,d(y,w)=d, d(x,w)=e,d(y,z)=f.$ We wish to show $|a-b|\leq c+d. $ We have $$(i).\quad b\leq c+e\leq c+(d+a)\implies b-a\leq c+d. $$ $$(ii).\quad a\leq c+f\leq c+(b+d)\implies a-b\leq c+d. $$
A: First prove this result, which is a very useful lemma in itself:
$$
\tag{*}\label{eq:ti}
|d(x, z) - d(y, z)| \leqslant d(x, y).
$$
Proof: this is equivalent to the conjunction of the inequalities
\begin{align*}
d(x, z) - d(y, z) & \leqslant d(x, y), \\
d(y, z) - d(x, z) & \leqslant d(x, y),
\end{align*}
each of which is a case of the Triangle Inequality (taking into account the symmetry of the function $d$).
Using \eqref{eq:ti} (twice), the symmetry of $d$ (again), and the Triangle Inequality for absolute values, we have
\begin{align*}
|d(x, y) - d(z, w)| & = |d(x, y) - d(z, y) + d(z, y) - d(z, w)| \\
& \leqslant |d(x, y) - d(z, y)| + |d(z, y) - d(z, w)| \\
& \leqslant d(x, z) + d(y, w).
\end{align*}
