Let $M = (X, d)$ be a D1107: Metric space.

Then
\begin{equation}
\forall \, x, y, z \in X : |d(x, z) - d(y, z)| \leq d(x, y)
\end{equation}

Result R708
on D1107: Metric space

Reverse triangle inequality for metric

Formulation 0

Let $M = (X, d)$ be a D1107: Metric space.

Then
\begin{equation}
\forall \, x, y, z \in X : |d(x, z) - d(y, z)| \leq d(x, y)
\end{equation}

Proofs

Let $M = (X, d)$ be a D1107: Metric space.

If $X$ is empty, the claim holds vacuously so assume that $X$ is nonempty and fix $x, y, z \in X$. By the triangle inequality metric axiom, we have $d(x, z) \leq d(x, y) + d(y, z)$. Subtracting $d(y, z)$ from both sides, this gives
\begin{equation}
d(x, z) - d(y, z)
\leq d(x, y)
\end{equation}
Next, again by triangle inequality, we have $d(y, z) \leq d(y, x) + d(x, z)$. Subtracting $d(x, z)$ from both sides and applying symmetry in $d(y, x) = d(x, y)$, this gives
\begin{equation}
- (d(x, z) - d(y, z))
= d(y, z) - d(x, z)
\leq d(x, y)
\end{equation}
Thus, by the definition of the D412: Absolute value function, we have the lower bound
\begin{equation}
|d(x, z) - d(y, z)|
\leq d(x, y)
\end{equation}
$\square$