Missing step to prove an inequality for bounded analytic function. The exercise is as follows
Exercise: Let $f : \overline{B(0,R)} \to \mathbb{C}$ be a holomorphism with $|f(z)| \leq M$, for some $M > 0$. Show that
$$\left| \frac{ f(z) - f(0)}{M^2 - \overline{f(0)}f(z)} \right| \leq \frac{|z|}{MR}.$$
My attempt: I defined the holomorphic function $g: B(0,1) \to \mathbb{C}$ 
$$ g(z) = M \frac{ f(Rz) - f(0)}{M^2 - \overline{f(0)}f(Rz)} $$
and the result would follow from Schwarz's Lemma if


*

*$g(B(0,1)) \subset B(0,1)$ 

*$g(0)=0$


The latter is easily seen, however I couldn't show the former.  
Cheers.
 A: It is easy to show that $g(0)=0.$
Note that $g(z)=\frac{\frac{f(Rz)}{M}-\frac{f(0)}{M}}{1-\frac{\overline{f(0)}}{M}\frac{f(Rz)}{M}}$, where $|\frac{f(Rz)}{M}|<1$ and $|\frac{f(0)}{M}|<1$. 
Let $\frac{f(Rz)}{M}=a+bi$, and $\frac{f(0)}{M}=c+di$, then:
$$g(z)=\frac{a+bi-c-di}{1-(a-bi)(c+di)}$$ and $$|g(z)|^2=\frac{(a-c)^2+(b-d)^2}{(1-ac-bd)^2+(cb-ad)^2}.$$
Since $(1-a^2-b^2)(1-c^2-d^2)>0$ (because of absolute values of $f/M<1$), doing simple calculation you can show that numerator of $|g(z)|^2$ is less then denomiantor, and so $|g(z)|^2<1$, and $g(B(0,1))$ belongs to $B(0,1).$
A: There is a theorem which could gives us some help here. It is in many books, for example, Rudin, Real and Complex Analysis, page 254.
12.4 Theorem: For any $\alpha \in B(0,1)$, define 
$$\varphi_\alpha(z) = \frac{z - \alpha}{1-\overline{\alpha}z}.$$
For fixed $\alpha \in B(0,1)$ the function $\varphi_\alpha$ is a one-to-one holomorphism which maps $B(0,1)$ onto $B(0,1)$, $ \partial B(0,1)$ onto $\partial B(0,1)$ and $\alpha$ to $0$. The inverse of $\varphi_\alpha$ is $\varphi_{-\alpha}$.
Sketch of the proof $\varphi_\alpha$ is holomorphic in the whole plane, except for a pole at $\frac{1}{\overline{\alpha}}$ which lies outside of $\overline{B(0,1)}$. Straightforward substitution shows that
$$\varphi_{-\alpha}(\varphi_\alpha(z))=z$$
Moreover, for $z \in \partial B(0,1)$ we have
$$| \varphi_\alpha(z) | = \frac{|z|}{|\overline{z}|}=1.$$
Hence by the Modulus Maximum Principle $|\varphi_\alpha(z)| < 1 $. The inverse gives us the surjective part.
Solution for the exercise: 
Using the manipulation given by @Jane, 
$$g(z) = \frac{ \frac{f(Rz)}{M} - \frac{f(0)}{M} }{ 1 - \frac{\overline{f(0)}}{M}\frac{f(Rz)}{M} }$$
we shall define $\alpha = \frac{f(0)}{M} \in B(0,1)$ and $h:B(0,1) \to B(0,1)$ with $h(z)=\frac{f(Rz)}{M}$.
Then by the theorem above,
$$g(z)= \varphi_\alpha ( h(z) ) $$
maps B(0,1) into B(0,1) and the results follows from Schwarz Lemma as already stated. 
