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Let $(X,\rho)$ be a metric space, and define $\sigma(x,y)=\min\{1,\rho(x,y)\}$. I have to prove that there are two positive constants $C_1,C_2$ such that

$C_1\rho(x,y)\leq\sigma(x,y)\leq C_2\rho(x,y)$.

Obviously we can take $C_2=1$, but how can we choose $C_1$?

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There are no such constants. Consider $X = \mathbb R$ and let $\rho$ be Euclidean metric. Now, $\rho(0,x) = |x|$ and $\sigma(0,x) = \min \{1,|x|\}$. Clearly, $ \lim\limits_{x\to+\infty}\frac{\sigma(0,x)}{\rho(0,x)}= 0\quad (1) $ but since $C_1 \rho(0,x)\leq \sigma(0,x)$ and $\rho(0,x)>0$ for $x>0$ we have: $ 0=\lim\limits_{x\to+\infty}\frac{\sigma(0,x)}{\rho(0,x)}\geq C_1 $ so $C_1 = 0$.

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Could it be that you've mixed up different definitions of equivalence of metrics? You're trying to prove strong equivalence for metrics that aren't strongly equivalent. They are, however, topologically equivalent: Since they coincide at small distances, the open sets they define coincide at small distances, so any open ball of one of them contains an open ball of the other and vice versa.