Let $R, C > 0$. Let f be a holomorphic function defined on $D(0, R)$ and such that f is bounded above by $C$. Prove that
$$|f'(z)| \leq \frac{R}{C}\cdot \frac{C^2 - |f(z)|^2}{R^2 - |z|^2}.$$
I am honestly just confused by this problem. My initial attempt was to reverse engineer the proof for Schwarz-Pick but it did not work out. Any thoughts?
$f$ is a holomorphic function from $D(0, R)$ into $ D(0, C)$, therefore $$ g(z) = \frac 1C f(Rz) $$ is defined for $z \in \Bbb D$ (the unit disk) with values in $\Bbb D$, and we can apply the Schwarz-Pick theorem to $g$: $$ |g'(z)| \le \frac{1-|g(z)|^2}{1-|z|^2} $$ for all $z \in \Bbb D$. Now we transform this result back to $f$: $$ f(z) = C g(\frac zR) \, , \quad f'(z) = \frac CR g'(\frac zR) \, , $$ so that $$ |f'(z)| = \frac CR | g'(\frac zR)| \le \frac CR \frac{1-|g(\frac zR)|^2}{1-|\frac zR|^2} = \ldots = \frac RC \frac{C^2 - |f(z)|^2}{R^2 - |z|^2} \, . $$