Complex Inequalities Let $\mathbb{H}=\{z\in \mathbb{C} | \ \Im(z)>0\}$ and $f:\mathbb{H} \to \mathbb{H}$ analytic. Prove that for every $z_1, z_2 \in \mathbb{H}$, it must happen that 
$$
\frac{|f(z_1)-f(z_2)|}{|f(z_1)-\overline{f(z_2)}|} \leq \frac{|z_1-z_2|}{|z_1-\overline{z_2}|}
$$
What I tried to do is fix a $z_2 \in \mathbb{H}$ and use the function 
$$
g(z)=\frac{f(z)-f(z_2)}{f(z)-\overline{f(z_2)}}
$$
Clearly $|g(z)|=1$ but I don´t see what´s the next step to do. Since this exercise appears in Ahlfors Complex Analysis after proving Schwarzs lemma, I think we must use it, but it don´t seems to apply for $g$, perhaps involving a Möbius map  can help? Please any help will be strongly appreciated.
 A: Fix $z_2\in\mathbb{H}$, and let
$$
g(w)=\frac{f(z)-f(z_2)}{f(z)-\overline{f(z_2)}}$$ where $w=\frac{z-z_2}{z-\overline{z_2}}$, so that $z=\frac{w\overline{z_2}-z_2}{w-1}$. Note that if $z_2=a+bi\in\mathbb{H}$ and $w=c+di\in\mathbb{D}$, then you can verify that 
$$
\Im z=\Im\frac{w\overline{z_2}-z_2}{w-1} =\frac{b(1-|w|^2)}{(c-1)^2+d^2}>0
$$
so $z\in\mathbb{H}$ and thus $f(z)$ is well-defined. Also, $|g(w)|<1$ which can be seen easily by noting that
$$
\left|\frac{f(z)-f(z_2)}{f(z)-\overline{f(z_2)}}\right|^2
=\frac{(\Re f(z)-\Re f(z_2))^2+(\Im f(z)-\Im f(z_2))^2}{(\Re f(z)-\Re f(z_2))^2+(\Im f(z)+\Im f(z_2))^2}
$$
while $(\Im f(z)-\Im f(z_2))^2<(\Im f(z))^2+(\Im f(z_2))^2<(\Im f(z)+\Im f(z_2))^2$ by positivity of $\Im f(z)$ and $\Im f(z_2)$. Finally, $g(0)=0$ (as $z=z_2$), so we apply the Schwarz lemma at the point $w=\frac{z_1-z_2}{z_1-\overline{z_2}}$ to see
$$
|g(w)|=\left|\frac{f(z_1)-f(z_2)}{f(z_1)-\overline{f(z_2)}}\right|\leq |w|=\left|\frac{z_1-z_2}{z_1-\overline{z_2}}\right|.
$$
