I'm trying to extend this result, i.e.,
Theorem: Let $(E, d)$ be a metric space and $n \in \mathbb N^*$. Let $d_n := \sqrt[n]{d}$. Then $d_n$ is a metric on $E$ and induces the same topology as $d$.
Could you have a check if my attempt is fine?
Proof: Clearly, $d_n (x,y)=0 \iff d(x, y)=0$. Let's prove that $d_n$ satisfies the triangle inequality. This follows directly from the following lemma.
Lemma: Let $a,b,c \ge 0$ such that $a \le b + c$. If $n \in \mathbb N^*$ then $\sqrt[n]{a} \le \sqrt[n]{b} + \sqrt[n]{c}$.
Proof: By Binomial theorem, we have $$ (\sqrt[n]{b} + \sqrt[n]{c})^{n} = \sum_{k=0}^{n} \binom{n}{k} b^{(n-k)/n} c^{k/n} = b+c + \sum_{k=1}^{n-1} \binom{n}{k} b^{(n-k)/n} c^{k/n}. $$ Then $(\sqrt[n]{b} + \sqrt[n]{c})^{n} \ge a$. The claim then follows.
To see that $d$ and $d_n$ induce the same topology on $E$, we notice that $d_n (x, y) < r \iff d (x, y) < r^n$. This completes the proof.
