Let $|\cdot|_1$ and $|\cdot|_2$ be two norms on a field $\mathbb F$. We call the two norms equivalent if every Cauchy-sequence with respect to $|\cdot|_1$ is also a Cauchy-sequence with respect to $|\cdot|_2$. Prove the following statement:
$|\cdot|_1\sim|\cdot|_2\quad\Leftrightarrow\quad\exists \alpha\in\mathbb{R}_{>0}: \forall x\in\mathbb F: |x|_1=|x|_1^\alpha.$
The direction "$\Leftarrow$" is straightforward: Let there an $\alpha$ with the property above, $(a_i)_{i\in\mathbb N}$ be a Cauchy sequence with respect to $|\cdot|_1$ and $\varepsilon_2\in\mathbb{R}_{>0}$ be arbitrary. Set $\varepsilon_1:=\varepsilon_2^\alpha$ and follow from the fact that $(a_i)_{i\in\mathbb N}$ is a Cauchy sequence, that there exists an $N\in\mathbb N$ such that $\forall n,m>N: |a_m-a_n|_1<\varepsilon_1.$ Since $|\cdot|_1\sim|\cdot|_2$ and the definition of $\varepsilon_1$ we get that $\forall n,m>N: |a_m-a_n|_2^\alpha<\varepsilon_1^\alpha$ which is equivalent to $\forall n,m>N: |a_m-a_n|_2<\varepsilon_1$ which means that $(a_i)_{i\in\mathbb N}$ is a Cauchy sequence with respect to $|\cdot|_2$.
For the other direction (which is probably harder) we may assume that the being a Cauchy sequence is the same property for both norms but I don't see how I can construct such an $\alpha$ from this fact.
Remark 1: This is an exercise number 5 on page 7 in the book "p-adic Numbers, p-adic Analysis and Zeta-Functions" (Second Edition) by Neal Koblitz.
Remark 2: The right side is the notion of norm-equivalence that I am familiar with, but in this book it is explicitly defined in the way from this post.