Möbius transformation mapping $\mathbb{D}$ into itself. I want to show that a Möbius transformation $f(z) = \frac{az + b}{cz + d}$ for $a,b,c,d \in \mathbb{C}$ maps the unit disk $\mathbb{D}$ into itself if and only if the coefficients satisfie
\begin{equation*}\label{ineq}
|\overline{b}d - \overline{a}c| + |ad - bc| \leq |d|^2 - |c|^2.
\end{equation*}

So far I have shown that if a Möbius transformation maps $\mathbb{D}$ into itself it satisfies the inequality.
Now my idea for the implication $\Leftarrow$:
I consider the inequality above and want to show that for
$|z| \leq 1 \implies$ $|f(z)| \leq 1$.
Instead of showing $|f(z)| \leq 1 $ I defined for $|d| > |c|$
\begin{equation*}
 v = \frac{b \overline{d} - a\overline{c}}{|d|^2 - |c|^2}, \; R = \frac{|ad - bc|}{|d|^2 - |c|^2}
\end{equation*}
and want to show $|f(z) - v|^2 \leq R^2$ since by the provided inequality
from $|f(z) - v|^2 \leq R^2$ follows $f(z) \in K_R(v) \subseteq \mathbb{D}$.
For the case $|d| = |c|$, $f$ wouldn't be a möbius transformation. 
Probelem: Now I have trouble showing the inequality above. I tried using the provided inequality
show that
\begin{equation*}
|f(z) - v|^2 = |f(z)|^2 - 2 \Re(f(z)\overline{v}) + |v|^2 \leq |f(z)|^2 - 2 \Re(f(z)\overline{v}) + 1 - 2R + R^2
\end{equation*}
but from then on I stuck by estimating
\begin{align*}
    |f(z)|^2 &= f(z) \overline{f(z)} = \frac{|a|^2 |z|^2 + 2\Re(az \overline{b}) + |b|^2}{|c|^2 |z|^2 + 2\Re(cz \overline{d}) + |d|^2} \\
    |v|^2 &= \frac{b \overline{d} - a\overline{c}}{|d|^2 - |c|^2}\frac{\overline{b} d - \overline{a}c}{|d|^2 - |c|^2} = \frac{|b|^2 |d|^2 - 2 \Re(b \overline{d} \overline{a}c) + |a|^2|c|^2}{(|d|^2 - |c|^2)^2} \\
    &2 \Re(f(z)v) = 2 \frac{1}{|d|^2 - |c|^2}\Re  \left( \frac{ab\overline{d}z - a^2 \overline{c}z  - ab\overline{c} + b^2 \overline{d}}{cz + d}\right)
\end{align*}
or showing that
\begin{align*}
|f(z)|^2 - 2 \Re(f(z)\overline{v}) + 1 - 2R &\leq 0 \\
\end{align*}
I made some estimates based on the fact that $|z| \leq 1$ but most of them lead nowhere.
I would appreciate it if someone can give me a hint or can tell that the idea doesn't work.

I thought to share my solution if someone is interested:
\begin{align*}
    |f(z) - w|^2 &= \left|\frac{az + b}{cz + d} - \frac{b \overline{d} - a\overline{c}}{|d|^2 - |c|^2}  \right|^2 \leq r^2 = \frac{|ad - bc|^2}{(|d|^2 - |c|^2)^2} \\
    &\iff \left|\frac{az + b}{cz + d}(|d|^2 - |c|^2) - b \overline{d} - a\overline{c}  \right|^2 \leq |ad - bc|^2 \\
    &\iff \left|(az + b)(|d|^2 - |c|^2) - (cz + d)b \overline{d} - a\overline{c}  \right|^2 \leq  |ad - bc|^2|cz + d|^2 \\
    &\iff \left| \overline{d}(azd + bd - czb -bd )  - \overline{c}(azc + bc - cza - da) \right|^2 \leq (|ad - bc| |cz + d|)^2 \\
    &\iff \left| \overline{d}z(ad - cb) - \overline{c}(bc - da) \right|^2 \leq (|ad - bc| |cz + d|)^2 \\
    &\iff \left| \overline{d}z + \overline{c} \right|^2 \leq |cz + d|^2 \\
    &\iff \left| \overline{d \overline{z} + c } \right|^2 \leq |cz + d|^2 \\
    &\iff \left| d \overline{z} + c \right|^2 \leq |cz + d|^2 
\end{align*}
and from $|z| \leq 1$ and $|d| \geq |c|$ we get
\begin{align*}
    |d \overline{z} + c|^2 &= |d|^2 |z|^2 + 2 \Re(d \overline{cz}) + |c|^2 \leq |cz + d|^2 = |c|^2|z|^2 + 2 \Re(cz \overline{d}) + |d|^2 \\
    &\iff |d|^2 |z|^2 + |c|^2 \leq |c|^2 |z|^2 + |d|^2 \\
    &\iff |d|^2(|z|^2 - 1) \leq |c|^2(|z|^2 - 1) \\
    &\iff |d|^2 \geq |c|^2 
\end{align*}
Therefore the inequality holds.
For $d = c$ $f$ is not a mobius transformation as I already stated above in my question.
