Let $f(z)=|z-1|/(|z|^p+1)^{1/p}$, is it true that $f(z.w) \le f(z)+f(w)$?

Question

Let $p \in \Bbb R$ with $p>1$ and $f_p : \Bbb C_{\ne 0} \rightarrow \Bbb R_{\ge 0}$ defined by $$f_p(z)=\frac{|z-1|}{(|z|^p+1)^{1/p}}$$ I'm trying to prove that $$f_p(z.w)\le f_p(z)+f_p(w)$$ I tried using "Desmos Graphing Calculator" to see if it was true for $z,w\in \Bbb R$ and it would seem to be the case (but there is no certainty). I tried proving it but end up with a headache so I began wondering if there is any smart trick to prove this. The reason I'm doing this is to prove the following function $$d_p(z,w)= \frac{|z-w|}{(|z|^p+|w|^p)^{1/p}}$$ Is a metric over $\Bbb C_{\ne0}$ with the following property $$d_p(z.w_1,z.w_2)=d_p(w_1,w_2) \quad \forall z,w_1,w_2 \in \Bbb C_{\ne 0}$$ Notice de importance of excluding de zero.