Complex analysis. Manipulation of conjugates, fractions and modulus. Let $a,b,c \in  \mathbb  C $  with $|b|<1$  and $z\neq \bar a$ and $$\left|\frac {z-a}{z-\bar a}\right| \le |b| $$
Show that,  $$|z| \le |a| \frac{1+|b|}{1-|b|}$$
This a revision question I'm doing on complex analysis and I'm getting about three quarters of the way through it but always end up with two extra terms which (as far as I can see) won't cancel to give the desired result.  This is what I have done so far:

We know that $z\bar z= |z|^2$ so if we square both sides $$|\frac {z-a}{z-\bar a}| \le |b|, $$ we obtain $$\frac {|(z-a)|^2}{|(z-\bar a)|^2} \le |b|^2 $$ or equivalently $$\frac {(z-a)(\bar z -\bar a)}{(z-\bar a)(\bar z- a)} \le |b|^2. $$ And through some manipulation, this becomes $$(|z|^2+|a|^2)(1-|b|^2) \le z\bar a +\bar za-|b|^2 za -|b|^2 \bar z \bar a. $$

I can't seem to get the desired result from this point, I've tried other methods but always end up at a point similar to this.
 A: (Note that if $a$ is real, then the left-hand side of the first inequality equals $1$, so the inequality can never be satisfied.) As long as $a$ is not real, the function $f(z)=(z-a)/(z-\bar a)$ is a linear fractional transformation; the first inequality says that the image of $z$ lies inside the circle of radius $|b|$ centered at $0$.
The inverse of $f(z)$ is $f^{-1}(w) = (a-\bar a w)/(1-w)$. If $|w|\le |b|$, then by two applications of the triangle inequality,
$$
|f^{-1}(w)| = |\bar a| \frac{|w-a/\bar a|}{|w-1|} \le |\bar a| \frac{|a/\bar a|+|w|}{1-|w|} = |a| \frac{1+|w|}{1-|w|} \le |a| \frac{1+|b|}{1-|b|},
$$
where the last inequality holds because $(1+t)/(1-t)$ is increasing for $0\le t<1$.
In other words, if $f(z)$ lands inside the circle of radius $|b|$ centered at the origin, then $z=f^{-1}(f(z))$ must satisfy the desired inequality.
A: From $\frac {|(z-a)|}{|(z-\bar a)|} \le |b|$, we have
$ {|(z-a)|}{} \le |b||(z-\bar a)|$
Apply triangle inequality $|p - q| \ge |p| - |q|$ to the LHS, we have
$ |z|- |a| \le |b||(z-\bar a)|$
On the RHS, replace the  $-\bar a$ by $ + {(-\bar a)}$, we have
$ |z|- |a| \le |b||z+ (-\bar a)|$
Let Apply triangle inequality $|p + q| \le |p| + |q|$ to the RHS, we have
$ |z| - |a| \le |b|(|z| + |(-\bar a)|)$
Since $|a| = |(-\bar a)|$ we have
$ |z| - |a| \le |b|(|z| + |a|)$
The result follows by re-arranging terms and making |z| as the subject. 
A: From the triangle inequality we have $|z|\leq|z-a|+|a|$, then 
\begin{align}
|z|-|a|\leq|z-a|\leq|b||z-\overline{a}| & =|bz-\overline{a}b|& \text{by properties of $|\cdot|$} \\
& \leq |bz|+|-\overline{a}b| & \text{by triangle inequality} \\
& = |b||z|+|\overline{a}||b|  & \text{by properties of $|\cdot|$}\\
& =|b||z|+|a||b|
\end{align}
Then 
\begin{align}
|z|-|b||z| & \leq |a|+|a||b| \\
(1-|b|)|z| & \leq |a|(1+|b|)
\end{align}
Therefore
$$|z|\leq |a|\dfrac{1+|b|}{1-|b|}.$$
