Let $g$ be an analytic function between two unit disks with $g(0)=a$, how to show that $|g'(a)|\le1-|a|^2$?
My idea is using Schwarz'Pick theorem. And by applying the theorem to $g\circ g(z)$, we have $|g'(a)g'(0)|\le\frac{1-|g\circ g(0)|^2}{1-|0|^2}=1-|g(a)|^2$.Thus $|g'(a)|\le\frac{1-|g(a)|^2}{|g'(0)|}$. I can't go further.
This is false. Noting that $(1-1/n)^{n} \to e^{-1}$ as $n \to \infty$ we can find $n$ such that $n(1-1/n)^{n-1}>4$. Fix such an integer $n$ and let $a=1-\frac 1 n$. Let $\phi(z)=\frac {z+a} {1+az}$ and $g(z)=\phi (z^{n})$. Since $g$ is a composition of two maps from the open unit disk into itself $g$ also has this property. Also $g(0)=\phi (0)=a$. Now $g'(z)= \frac {(1+az^{n})nz^{n-1}-(z^{n}+a)naz^{n-1}} {(1+az^{n})^{2}}$. Hence $g'(a)= g'(z)= \frac {(1+a^{n+1})na^{n-1}-(a^{n}+a)na^{n}} {(1+a^{n+1})^{2}}$. This gives $g'(a)=\frac {na^{n-1}(1-a^{2})} {(1+a^{n+1})^{2}}$, If $|g'(a)| \leq (1-a^{2})$ then we get ${na^{n-1}}\leq (1+a^{n+1})^{2} \leq 2^{2}=4$. This contradicts the choice of $a$.