I'm solving a problem sheet in complex analysis. We defined the pseudo-hyperbolic metric on $\mathbb{D}=\{z\in\mathbb{C}\mid |z| < 1\}$ as $d_{ph}(z,w)=\left|\frac{z-w}{1-\overline{w}z}\right|$ and the pseudo-hyperbolic disc as the set $D_{ph}(a,r)=\{d_{ph}(z,a)<r\}$. Why does for a holomorphic map $\varphi\colon\mathbb{D}\to\mathbb{D}$ the following inclusion hold $$\varphi(D_{ph}(a,r))\subseteq D_{ph}(\varphi(a),r)?$$ I tried computing the conditions but that lead me nowhere. Is there a geometric interpretation of the given inclusion and is it connected to the fact that holomorphic maps on $\mathbb{D}$ are contractions with regard to the metric $d_{ph}$? I computed that the euclidean centre and radius of $D_{ph}(a,r)$ are $\frac{(1-r^2)a}{1-|a|^2r^2}$ and $\frac{(1-|a|^2)r}{1-|a|^2r^2}$ respectively.
Update: I've been thinking thinking for the whole day. From my TA i got the hint: Schwarz-Pick lemma but it doesn't help me.
Yes we can use Schwarz-Pick Lemma. In shorts Schwarz-Pick Lemma states that holomorphic functions are distance decreasing. So lets begin the proof: Let $z\in \varphi(D_{ph}(a,r))\implies \exists \alpha \in D_{ph}(a,r)$ such that $\varphi(\alpha)=z$. Since $\alpha \in D_{ph}(a,r) \implies d_{ph}(\alpha,a)= \frac{\alpha-a}{1-\bar{a}\alpha}< r$, as Schwarz-Pick lemma is distance decreasing $\implies d_{ph}(\varphi(\alpha),\varphi(a))<d_{ph}(\alpha,a)<r $ and $\implies z\in D_{ph}(\varphi(a),r)$ and hence the inequality.