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

$d$ 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.
