I am stuck on a relation composition problem from book [1]. The context is: Suppose $r \in \mathbb{R}^+$ and $s \in \mathbb{R}^+$. Let $D_r = \{(x, y) \in \mathbb{R}^2 \mid |x-y| < r\}$, and $D_s = \{(x, y) \in \mathbb{R}^2 \mid |x-y| < s\}$. The problem is only: what is $D_r \circ D_s$?
From the definition of composition of relations, \begin{align} D_r \circ D_s &= \{(x, z) \in \mathbb{R}^2 \mid \exists y \in \mathbb{R}(xD_sy\ \text{and}\ yD_rz)\} \\ &= \{(x, z) \in \mathbb{R}^2 \mid \exists y \in \mathbb{R}(|x-y| < s\ \text{and}\ |y-z| < r)\}. \end{align} The book hints to use the triangle inequality $\forall x \in \mathbb{R}\forall y \in \mathbb{R}(|x + y| \leq |x| + |y|)$. My suspicion is that the intended answer is \begin{equation} D_r \circ D_s = \{(x, z) \in \mathbb{R}^2 \mid |x - z| < s + r\}. \end{equation} (If it isn't, then I fail to see how the triangle inequality relates.) But for this to be the case it must be that \begin{equation} \exists y \in \mathbb{R} (|x-y| < s\ \text{and}\ |y-z| < r) \leftrightarrow |x - z| < s + r \end{equation} given the above assumptions on $r, s, x, y$. Proving ($\rightarrow$) is straightforward: suppose the LHS is true, choose some $y \in \mathbb{R}$ such that $|x - y| < s$ and $|y - z| < r$, then $|x - y| + |y - z| < s + r$, but $|x - z| = |(x - y) + (y - z)| \leq |x - y| + |y - z|$ (triangle inequality), so $|x - z| < s + r$.
How to prove ($\leftarrow$) is not clear to me. After assuming the RHS, is there a particular value of $y \in \mathbb{R}$ that I can choose to make the LHS true?
[1] Velleman, Daniel. How to Prove It: A Structured Approach 2e, Ch. 4 Sec. 3 Prob 6.