Let $\alpha \in \mathbb C$ such that $|\alpha| \in (0,1)$. Prove that if $z\in \mathbb C$ is such that $|z| \le r < 1$ then :
$$ \frac{\alpha+|\alpha| z}{\alpha(1-\overline{\alpha} z)}\le \frac{1+r}{1-r} $$
I tried to use the Maximum Modulus Principle. Let $z\in \partial B(0,r) $ i.e $z=re^{i\theta}$ for some $\theta \in \mathbb R$. It's enough to looking for the maximum of:
$$ \frac{|\alpha+|\alpha| re^{i\theta}|}{|\alpha(1-\overline{\alpha} re^{i\theta})|}= \frac{|1+\frac{|\alpha|}{\alpha} re^{i\theta}|}{|1-\overline{\alpha} re^{i\theta}|}$$
I don't know how to proceed :/
This does not require anything other than the triangle inequality. Estimate the numerator from above: $$ |\alpha+|\alpha| z|\le |\alpha|(1+r) $$ and the denominator from below: $$ |\alpha(1-\overline{\alpha} z)| \ge |\alpha|(1-|\alpha|r) \ge |\alpha|(1-r) $$