Proving $\frac{|f'(z)|}{1-|f(z)|^2}\leq\frac{1}{1-|z|^2}$ from the Schwarz--Pick Lemma without using other inequality

Question

Question: Let $f$ be analytic in $\mathbb{D}$. Show that for $|f(z)|\leq1$ and $|z|<1$, we get $$\frac{|f'(z)|}{1-|f(z)|^2}\leq\frac{1}{1-|z|^2}$$

This is one of the two inequalities in the Schwarz--Pick Lemma. Every proof that I am seeing of this lemma is using the other inequality from the Schwarz--Pick Lemma. That is, given the same assumptions with $z_1, z_2\in\mathbb{D}$, we have $$\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right|\leq\left|\frac{z_1-z_2}{1-\overline{z_1}z_2}\right|$$ I was hoping to find a proof that was not reliant on this inequality, but rather stood alone (again, just trying to prove the inequality with the derivative part in it).

Any help is appreciated! Thank you.

Precise Statement of Schwarz-Pick Lemma I am referencing: Let $f:\mathbb{D}\rightarrow\mathbb{D}$ be analytic. Then, for all $z,z_1,z_2\in\mathbb{D}$, we have $$\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right|\leq\left|\frac{z_1-z_2}{1-\overline{z_1}z_2}\right|$$ and $$\frac{|f'(z)|}{1-|f(z)|^2}\leq\frac{1}{1-|z|^2}$$

Answer 1

Schwarz-Pick: If $f$ is holomorphic on the unit disc with $|f(z)| \leq 1$ for all $z \in D(0,1)$, then $\frac{|f'(z)|}{1-|f(z)|^2}\leq\frac{1}{1-|z|^2}$. This can be ascertain as follows:

Fix $c \in D(0,1)$ and set $\phi_{c}(z) = \frac{z - c}{1 - \overline{c}z}$. Show that $\phi_{c}$ is a bijection from $D(0,1)$ to itself and the inverse is given by $\phi_{-c}$. Now, define $g(z) = \phi_{f(z)} \circ f \circ \phi_{-z}$. Show that $g(z)$ satisfies the assumption of the Schwarz Lemma. The desired inequality drops out from the inequality $g'(0) \leq 1$ (arising from the Schwarz Lemma).