Are $d$ and $d' = \min (d(x,y), 1)$ strongly equivalent?

Question

Let $(X, d)$ be a metric space. Are $d$ and $d' = \min (d(x,y), 1)$ strongly equivalent?

From the definition, it is clear that $d'(x,y)= \min (d(x,y), 1)\leq d(x,y). $ Here I get $\beta=1$. I tried to prove the reverse inequality for some $\alpha$, I couldn't. I think these two are not equivalent.

Answer 1

$d$ and $d_1$ are strongly equivalent $\iff$ $d$ is bounded.

$\implies$: If $d$ and $d_1$ are strongly equivalent then for some $\alpha > 0$ and all $x, y \in X$ $$ d(x, y) \le \alpha d_1(x, y) \le \alpha \, , $$ so that $d$ is bounded.

$\impliedby$: If $d(x, y) \le K$ for some $K > 0$ and all $x, y \in X$, then $$ d_1(x, y) \le d(x, y) \le \max(1, K) d_1(x, y) $$ so that the metrics are strongly equivalent. The last inequality holds because both $$ \begin{aligned} d(x, y) &\le K \cdot 1 \le \max(1, K) \cdot 1 \\ d(x, y) &= 1 \cdot d(x, y) \le \max(1, K) \cdot d(x, y) \end{aligned} $$ are true.