Two norms $\Vert -\Vert _1 $, $\Vert -\Vert _2$ are equivalent if:
for two constants $a,b$ and $x$ from $V$ a vector space over a field it holds that: $$a\Vert x\Vert _1\leqslant \Vert x\Vert _2\leqslant b\Vert x\Vert _1.$$
This is a equivalence relation because:
$$a\Vert x\Vert _1\leqslant \Vert x\Vert _2\leqslant b\Vert x\Vert _1$$ and $$c\Vert x\Vert _2\leqslant \Vert x\Vert _3 \leqslant d\Vert x\Vert _2$$
it follows that there are constants such that (transitivity): $$e\Vert x\Vert _1 \leqslant \Vert x\Vert _3 \leqslant f\Vert x\Vert _1$$
Reflexivity: $$a\Vert x\Vert _1\leqslant \Vert x\Vert _1\leqslant b\Vert x\Vert _1$$ with $a,b = 1$; this is true.
Symmetry: $$a\Vert x\Vert _1 \leqslant \Vert x\Vert _2 \leqslant b\Vert x\Vert _1$$
if we take: $$-\frac{1}{b}\Vert x\Vert _2\leqslant \Vert x \Vert_1 \leqslant \frac{-1}{a} \Vert x\Vert _2.$$
Is this a valid proof that the equivalence of two norms is truly a equivalence relationship?
Attempt 2:
Symmetry:
$$a\Vert x\Vert _1 \leqslant \Vert x\Vert _2 \leqslant b\Vert x\Vert _1$$
$\Rightarrow $: $$\frac{1}{b} \Vert x\Vert _2 \leqslant \Vert x\Vert _1 \leqslant \frac{1}{a}\Vert x\Vert _2.$$
Transitivity: given
$$a\Vert x\Vert _1\leqslant \Vert x\Vert _2\leqslant b\Vert x\Vert _1$$ and $$c\Vert x\Vert _2\leqslant \Vert x\Vert _3 \leqslant d\Vert x\Vert _2$$
$\Rightarrow$ : $$ac \Vert x\Vert _1 \leqslant c\Vert x\Vert _2\leqslant \Vert x\Vert _3 \leqslant d\Vert x\Vert _2\leqslant db\Vert x\Vert _1.$$
$ \newcommand{\norm}[2]{\left \Vert #1 \right \Vert_{#2}} \newcommand{\N}{\mathcal{N}(V)} \newcommand{\a}{\alpha} \newcommand{\b}{\beta} \newcommand{\g}{\gamma} \newcommand{\l}[1]{\stackrel{(#1)}{\le}} $ I feel like your proof could be better phrased to make it clearer what the conditions imply and require, where they come from, and your understanding thereof, at least for your scratch work. I come away from your proof not fully sure you even know what needs to be verified. (Granted, this could entirely be a language barrier problem.) I will try to rewrite the proof in such a way to make these notions and your writing clearer, if just for future readers if not to enlighten you.
First though, some more specific criticisms:
Defining the Relation: So we define the relation as so: for norms $\norm \cdot \a, \norm \cdot \b$ on $V$ a vector space over an ordered field $F$,
$$\norm \cdot \a \sim \norm \cdot \b \iff (\forall x \in V)(\exists a,b \in F^+) \left(a \norm x \a \le \norm x \b \le b \norm x \a\right)$$
To show $\sim$ is an equivalence, we need to verify the three necessary conditions. Note that $\sim$ is a relation on the set $\N$ of norms $\norm \cdot \a : V \to F$. Once verified, we will say that $\sim$ denotes the equivalence of norms.
Reflexivity: The requirement here is that $\norm \cdot \a \sim \norm \cdot \a$ for every $\norm \cdot \a \in \N$. This holds by taking $a=b=1$, trivially.
Symmetry: Suppose $\norm \cdot \a \sim \norm \cdot \b$. Then to show symmetry we will want to show $\norm \cdot \b \sim \norm \cdot \a$. So, we know that
$$(\forall x \in V)(\exists a,b \in F^+)\left(a \norm x \a \le \norm x \b \le b \norm x \a\right) \tag 1$$
What we want to show is the similar condition
$$(\forall x \in V)(\exists a',b' \in F^+)\left(a' \norm x \b \le \norm x \a \le b' \norm x \b\right)$$
Consider the inequality in $(1)$. Divide by $a$ to obtain $(2a)$, and by $b$ to obtain $(2b)$. Note that inequality is maintained since $a,b > 0$.
\begin{align*} &\norm x \a \le \frac 1 a \norm x \b \le \frac b a \norm x \a \tag{2a} \\ \frac a b &\norm x \a \le \frac 1 b \norm x \b \le \norm x \a \tag{2b} \end{align*}
Combine the two about $\norm x \a$ and we have the chain of inequalities $(3)$, with the boxed portion being most relevant:
$$\frac a b \norm x \a \le \boxed{\frac 1 b \norm x \b \le \norm x \a \le \frac 1 a \norm x \b} \le \frac b a \norm x \a \tag{3}$$
Since $a,b > 0$, then $1/a,1/b>0$ as well. Thus, take $a' = 1/a$ and $b' = 1/b$ to obtain symmetry!
Transitivity: So, suppose $\norm \cdot \a \sim \norm \cdot \b$ and $\norm \cdot \b \sim \norm \cdot \g$. We want to show $\norm \cdot \a \sim \norm \cdot \g$. So, what we know is that
\begin{align*} (\forall x \in V)(\exists a,b \in F^+) &\left(a \norm x \a \le \norm x \b \le b \norm x \a\right) \tag 4\\ (\forall x \in V)(\exists c,d \in F^+) &\left(c \norm x \b \le \norm x \g \le d \norm x \b\right) \tag 5 \end{align*}
where the first known relation is in $(4)$, and the second is in $(5)$. On the basis of these, we want to demonstrate $(6)$:
$$(\forall x \in V)(\exists e,f \in F^+) \left(e \norm x \a \le \norm x \g \le f \norm x \a\right) \tag 6$$
We now pass around between the inequalities in $(4)$ and $(5)$, using $\norm x \g$ as a starting point, and constructing inequalities each way. We obtain the following (where the numbers above each inequality show which one implies the inequality):
$$ a c \norm x \a \l 4 c \norm x \b \l 5 \norm x \g \l 5 d \norm x \b \l 4 bd \norm x \a $$
For emphasis, focus on the first, third, and fifth terms:
$$ a c \norm x \a \le \norm x \g \le bd \norm x \a $$
Thus, take $e = ac$ and $f=bd$ in $(6)$ to obtain transitivity! And, with all three properties verified, $\sim$ is an equivalence relation!
...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.