Show that if $(X,p)$ is complete, then so is $(X,d)$, where $\frac{3}{2022}p(x,y) \le d(x,y) \le \min\{1,p(x,y)\}$, for any $x,y \in X$. Let $d$ and $p$ be two metrics on $X$ such that
$$\frac{3}{2022}p(x,y) \le d(x,y) \le \min\{1,p(x,y)\},$$
for all $x,y \in X$. Show that if $(X,p)$ is complete, then so is $(X,d)$.
What I think:
Let $(x_n)$ be any Cauchy sequence in $(X,d)$. I know that the goal is to show that $(x_n)$ is convergent in $(X,d)$. It can be done by showing that $(x_n)$ is a Cauchy sequence in $(X,p)$, since $(X,p)$ is complete. But, I didn't know yet how to apply to there.
Since $(x_n)$ is a Cauchy sequence in $(X,d)$, then for any $\epsilon>0$, there is $N \in \Bbb N$ such that for all $m,n \ge N$, we have $d(x_n,x_m)<\epsilon$. I got stuck when I want to show that $p(x_n,x_m)<\epsilon$. Is it true that by hypothesis, $d(x_n,x_m)<p(x_n,x_m)<\epsilon$? If yes, how to approach it?
Any helps? Thanks in advanced.
 A: Let $(x_n) $ $d$- cauchy.
Then $p(x_m, x_n) \le \frac{2022}{3}d(x_m, x_n)\to 0 \text{ as } m, n\to \infty$
Hence $(x_n) $ $p$-cauchy.
$(X, p) $ complete imply $x_n\to x$ in $(X, p) $
$d(x_n, x) \le \min\{1,p(x_n,x)\}\to  0 \text{ as } n\to \infty$
Hence $(x_n) \to x $ in $(X, d) $

"$\min$ " function  is continuous. $(x_n) \to x , (y_n) \to y$ implies $\min\{x_n,y_n\}\to \min\{x,y\}$
$(x_n) =(1, 1,1,\ldots) \to 1$
$(y_n) =(p(x_n, x)) \to 0$
Hence $\min\{1,p(x_n,x)\}\to \min\{1,0\}=0$
A: Let $(x_n)$ be any Cauchy sequence in $(X,d)$, i.e., for all $\epsilon>0$, there exists $N \in \Bbb N$ such that for any $n,m\ge N$, we have $d(x_n,x_m)<\frac{3}{2022}\epsilon$. By hypothesis of $d$, we have
\begin{align*}
p(x_n,x_m) &\le \frac{2022}{3} d(x_n,x_m)\\
&< \frac{2022}{3} \cdot \frac{3}{2022} \epsilon \\
&= \epsilon,
\end{align*}
for every $n,m \ge N$.
Hence, $(x_n)$ is a Cauchy sequence in the complete metric space $(X,p)$.
Thus, $x_n \to x$ for some $x \in X$. In other words, for any $s>0$, there exists $M \in \Bbb N$ such that for all $n\ge M$, we have $p(x_n,x)<s$.
Now, again, by hypothesis of $d$, we have $x_n \to x$ in $(X,d)$ as well. Indeed, for any $s>0$, there exists $M \in \Bbb N$ such that for all $n\ge M$, we have:
\begin{align*}
d(x_n,x) &\le \min\{1,p(x_n,x)\} \\
&\le p(x_n,x) \\
&< s.
\end{align*}
Hence, every Cauchy sequence in $(X,d)$ is convergent. Therefore, $(X,d)$ is complete, as desired.
Hope this helps.
