The following problem is from 'Applied Complex Variables for Scientists and Engineers' by Yue Kuen Kwok.
Suppose f(z) is analytic on and inside the unit circle $|z|=1$, and that Re $f(z) > 0,$ with $f(0) = \alpha > 0$.
Show that
$$\left| \frac{f(z)-\alpha}{f(z)+\alpha} \right| \leq |z| \hspace{5mm} \text{ and } \hspace{5mm} |f'(0)| \leq 2 \alpha.$$
Attempt: for the first inequality, if we write $f(z)$ as $u(x,y)+iv(x,y)$, we have $$LHS = \frac{(u-\alpha)^2 +v^2}{(u+\alpha)^2 +v^2} \leq \frac{u^2 - 2u \alpha + \alpha^2 +v^2}{u^2+2u\alpha + \alpha^2 + v^2} \leq \frac{u^2 + 2u \alpha + \alpha^2 +v^2}{u^2+2u\alpha + \alpha^2 + v^2} = 1,$$ meaning that is easy enough to show the LHS is less than or equal to one, but I cannot seem to make any modifications to show it is less than $|z|$, or think of any other way to do it... if anyone has any ideas I'd be very grateful. I suspect the maximum modulus principle applies somehow...
Note: this is from a chapter which covers Cauchy int. thm., max mod principle, and mean value theorem only, and so Schwarz/Harnack/other inequalities should not be employed.
Step 1. Set $$ g(z)=\frac{f(z)-a}{f(z)+a}, $$ and show that $g$ is analytic in the unit disk, $|g(z)|\le 1$ and $g(0)=0$.
Step 2. Apply Schwarz Lemma to $g$, to obtain that $$|g(z)|\le |z| \quad \text{and}\quad |g'(0)|\le 1. $$
Step 3. Translate the inequalities above for $g$, as inequalities for $f$.