Prove that a set A is open with respect to metric norm dp if and only if it is open with dq

Consider two metrics $$d_{p}=(\sum\limits_{k=1}^n |x_{k}-y_{k}|^p)^{1/p}$$ and $$d_{q}=(\sum\limits_{k=1}^n |x_{k}-y_{k}|^q)^{1/q}$$

Prove that a non-empty subset $$A \subset \mathbb R ^n$$ is open with respect to $$d_{p}$$ iff it is open with respect to $$d_{p}$$

Attempt

So $$A$$ is open in both those metrics if $$d_{p}$$ and $$d_{q}$$ are equivalent.

Then I have to prove that $$\alpha d_p(x,y) \leq d_q(x,y) \leq \beta d_p(x,y)$$ for $$\alpha, \beta$$ positive.

How do I start to prove this?

If $$p define $$A_i=\frac{|x_i-y_i|^{1/p}}{\sum_{j=1}^n |x_j-y_j|^{1/p}}\le 1$$ so $$f(x)=\sum_{i=1}^nA_i^x$$ has negative derivative and thus $$n\ge f(0)\ge f(p/q)\ge f(1)=1$$ implies $$nd_p(x,y)^{p^2}\ge d_q(x,y)^{q^2} \ge d_p(x,y)^{p^2}$$. Now, if $$A$$ is open under $$d_p$$, for any $$x \in A$$ there exists $$r$$ s.t. $$d_p(x,y). Choosing $$R=r^{p^2/q^2}$$ shows that $$d_q(x,y) implies $$d_p(x,y)\le d_q(x,y)^{q^2/p^2} so $$y\in A$$ and $$A$$ is open under $$d_q$$. The converse follows similarly.