A problem of Schwarz-Pick theorem in complex analysis Set $f \in H(B(0,1)),f(B(0,1)) \subset B(0,1)$，prove that
$$|f(z)-\frac{f(0)(1-|z|^2)}{1-|f(0)|^2||z|^2}| \le \frac{|z|(1-|f(0)|^2)}{1-|z|^2|f(0)|^2}$$
I think this problem is supposed to be proved by Schwarz-Pick theorem, I tried hard but failed to build a function to prove it. I even doubted that this problem is wrong, so I want to whether this problem can be proved or be falsified, and how.
Any help would be appreciated. 
 A: We may assume that $p=f(0)$ is real positive without loss of generality.
We consider a Mobius transformation $$\varphi (z)=\frac{z+p}{1+pz},\quad \varphi : B(0,1) \to B(0,1).$$
Let $C_r=\{z: |z|=r\}$, $B_r=B(0,r)$ for $r$ with $0<r<1.$ 
Note that $\varphi (C_r),$ the boundary of a disc $\varphi (B_r),$ is a circle symmetric with respect to the real axis.   
Since $\varphi $ is bijective, $\varphi (0)=f(0)$ and $f(B)\subset \varphi (B)$ we have 
$$f(B_r)\subset \varphi (B_r).\tag{1}$$
This is a result of Lindeloef' subordination principle. However, in this case we can get it easily by Schwarz lemma.  
And now let $$q=\frac{p(1-r^2)}{1-p^2r^2}.$$ The circle 
$\varphi (C_r)$ intersects with the real axis in two points $x=\frac{r+p}{1+pr}$ and $\frac{-r+p}{1-pr}.$  By the symmetry of $\varphi (C_r)$,
$$\frac{\,1\,}{2}\left(\frac{r+p}{1+pr}+\frac{-r+p}{1-pr}\right)=q$$
is $x$-coordinate of the center of $\varphi (C_r)$ and $$
\frac{\,1\,}{2}\left(\frac{r+p}{1+pr}-\frac{-r+p}{1-pr}\right)=\frac{r(1-p^2)}{1-p^2r^2}$$
is the radius.
Since $f(C_r)\subset \varphi (B_r)$ by $(1)$, we have$$
|f(re^{i\theta })-q|\le \frac{r(1-p^2)}{1-p^2r^2},\quad (0\le \theta <2\pi).$$
Thus we have done.
