Let $\mathbb{C},\;\mathbb{D}$ denote the complex plane and the unit disk in $\mathbb{C},$ respectively.
Suppose that $z,\;a\in\mathbb{D}.$ Then the Moebius transfromation w.r.t. $a\in\mathbb{D}$ satisfies that
\begin{equation}\label{MOB}
\left|\frac{z-a}{1-\overline{a}z}\right| \le \delta < 1
\end{equation}
Since $|a|,\delta<1,$ it holds
$$(\ref{MOB})
\iff |z-a|^{2} \le \delta^{2}|1 - \overline{a}z|^{2}
$$$$
\iff (z-a)(\overline{z}-\overline{a}) \le \delta^{2} (1-\overline{a}z)(1-a\overline{z})
$$$$
\iff |z|^{2} + |a|^{2} - (z\overline{a}+\overline{z}a) \le \delta^{2} (1+|a|^{2}|z|^{2} -(z\overline{a}+\overline{z}a))
$$$$
\iff |z|^{2} - \frac{1-\delta^{2}}{1-\delta^{2}|a|^{2}}(z\overline{a}+\overline{z}a) \le \frac{\delta^{2} - |a|^{2}}{1-\delta^{2}|a|^{2}}
$$$$
\iff \left(z - \frac{1-\delta^{2}}{1-\delta^{2}|a|^{2}}a\right)\left(\overline{z} - \frac{1-\delta^{2}}{1-\delta^{2}|a|^{2}}\overline{a}\right)
\le \frac{\delta^{2} - |a|^{2}}{1-\delta^{2}|a|^{2}} + \frac{(1-\delta^{2})^{2}|a|^{2}}{(1-\delta^{2}|a|^{2})^{2}}
$$$$
\iff \left|z - \frac{1-\delta^{2}}{1-\delta^{2}|a|^{2}}a\right|^{2}
\le \frac{(1-|a|^{2})^{2}\delta^{2}}{(1-\delta^{2}|a|^{2})^{2}}
$$$$
\iff \left|z - \frac{1-\delta^{2}}{1-\delta^{2}|a|^{2}}a\right|
\le \frac{(1-|a|^{2})\delta}{1-\delta^{2}|a|^{2}}.
$$ It follows that
$$
\left|\frac{z-a}{1-\overline{a}z}\right| \le \delta \iff
\left|z - \frac{(1-\delta^{2})a}{1-\delta^{2}|a|^{2}}\right|
\le \frac{(1-|a|^{2})\delta}{1-\delta^{2}|a|^{2}}.
$$
By the triangle inequality, it follows
$$
|z|-\frac{(1-\delta^{2})|a|}{1-\delta^{2}|a|^{2}}
\le
\left|z - \frac{(1-\delta^{2})a}{1-\delta^{2}|a|^{2}}\right|
\le \frac{(1-|a|^{2})\delta}{1-\delta^{2}|a|^{2}}
$$
and we have
$$|z| \le \frac{(1-\delta^{2})|a|}{1-\delta^{2}|a|^{2}} + \frac{(1-|a|^{2})\delta}{1-\delta^{2}|a|^{2}} = \frac{\delta+|a|}{1+|a|\delta}.
$$
