Schwarz Inequality of function from upper half plane to disc

So I've been working on this problem and I have everything nailed down (I think) except for the very end. In particular I get a bound, but I can't seem to reduce it down to the one the question is looking for. The question is:

Suppose $f$ maps the $\{z\in\mathbb{C}:\textrm{Im}(z)>0\}$ onto the unit disc $\mathbb{D}.$ Using Schwarz's Lemma prove $$|f(z)|^2+|f'(z)|\leq 1$$ for all $z\in\left\{z\in\mathbb{C}:\textrm{Im}(z)\geq\frac{1}{2}\right\}.$

So the first thing I did was use $g(z)=i\frac{1-z}{1+z}$ to map the disc to the upper half plane, and so $g\circ f$ is an automorphism of the disc. From here I chose $a\in\mathbb{D},$ and let $b=f(g(a)),$ and constructed the maps $\Psi_a$ and $\Psi_b,$ where $$\Psi_w(z)=\frac{w-z}{1-\overline{w}z}.$$ Hence, the composition given by $F(z)=(\Psi_a\circ f\circ g\circ\Psi_b)(z)$ meets the criterion set by Schwarz's Lemma. So we have $$|F'(0)|\leq 1 \Rightarrow |f'(g(a))g'(a)|\leq\frac{1-|b|^2}{1-|a|^2} \Rightarrow |f'(g(a))|\leq \frac{1-|b|^2}{1-|a|^2}\frac{|(1+a)^2|}{2}.$$ Now I want to let $g(a)=z,$ and so $b=f(z),$ (but I should probably not until the end?), which gives me $$|f'(z)|\leq \frac{1-|f(z)|^2}{1-|a|^2}\frac{|(1+a)^2|}{2},$$ and this is what I want provided I can show that $$\frac{|(1+a)^2|}{2(1-|a|^2)}\leq 1.$$ That inequality is the bit I have been having issues with. I think I am just missing some clever application of the fact $\textrm{Im}(z)>\frac{1}{2},$ but I keep trying and nada. Mostly I have been trying to rewrite the modulus as the product of the number and its conjugate..

Any push in the right direction is greatly appreciated.