Prove $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent iff there are positive scalars $m$ and $M$ s.t. every $v$, $m\|v\|_1 \leq \|v\|_2 \leq M\|v\|_1$ Exercise
We say two norms are equivalent iff its induced metrics are equivalent as well.
Prove the following statements are equivalent:
(a) $\|\cdot\|_{1}$ and $\|\cdot\|_{2}$ are equivalent.
(b) There are $m\in\mathbb{R}_{>0}$ and $M\in\mathbb{R}_{>0}$ such that, for every  $v$ it holds that
\begin{align*}
 m\|v\|_{1} \leq \|v\|_{2} \leq M\|v\|_{1}.
\end{align*}
My attempt
Let us prove the implication $(b)\Rightarrow(a)$ first.  According to the hypothesis, for every $x\in X$ and every $r > 0$, there exists $s = r/M > 0$ such that for every $y\in X$
\begin{align*}
\|x - y\|_{1} < s \Rightarrow \|x - y\|_{2} < r.
\end{align*}
Similarly, for every $x\in X$ and every $r > 0$, there exists $t = mr > 0$ such that for every $y\in X$
\begin{align*}
\|x - y\|_{2} < t \Rightarrow \|x - y\|_{1} < r.
\end{align*}
In this way, we have just proven that $\|\cdot\|_{1}\sim\|\cdot\|_{2}$.
However I am not able to prove the implication $(a)\Rightarrow(b)$.
Can somebody help me with this?
EDIT
We say two metrics $d$ and $d'$ defined over a non-empty set $X$ are equivalent iff for every $x\in X$ and every $r > 0$, there are $s > 0$ and $t > 0$ such that
\begin{align*}
\begin{cases}
\{y\in X : d'(x,y) < s\} \subset \{y\in Y : d(x,y) < r\},\\\\
\{y\in X : d(x,y) < t\} \subset \{y\in Y : d'(x,y) < r\}.
\end{cases}
\end{align*}

Suppose that (a) holds. By your definition of equivalence, with $r=1$ there exists $s>0$, $t>0$ such that 1. if $\| x\|_2 <s$ then $\|x\|_1<1$ and 2. if $\| x\|_1 <t$ then $\|x\|_2<1$.
Let $x\in X$ be arbitrary and let $\omega = sx/2\|x\|_2$. Then $$\|\omega \|_2=\frac 
 s 2 <s$$ so $\| \omega\|_1<1$. This implies that $s \frac{\|x\|_1}{2\|x\|_2}<1 $ so $$\| x\|_1 \leqslant \frac 2 s \| x\|_2. $$ Similarly by considering $\tilde \omega = tx/2\|x\|_1$ you obtain the other inequality.
