If $|z| < 1$, prove that $\Re \left(\frac{1}{1 - z} \right) > \frac{1}{2}$. 
If $|z| < 1$, prove that $\Re \left(\frac{1}{1 - z} \right) > \frac{1}{2}$.

My attempt:
Consider $\frac{1}{1 - z}$. Let $z = x + iy$, we know that $|z| < 1 \implies x, y < 1$. $$\frac{1}{1 - z} = \frac{1}{1 - x - iy} = \frac{1 - x + iy}{(1 - x)^2 + y^2}.$$
$$\Re \left(\frac{1}{1 - z} \right) = \frac{1 - x}{(1 - x)^2 + y^2}.$$
I got $$\frac{1}{(x-1)^2 + y^2} > \frac{1}{2}.$$
How do I manage the numerator? Can you help me? I welcome the alternative approaches.

You need to use more than just $x,y<1$ for example with $x=y=\frac{3}{4}$ you obtain $\Re\left(\frac{1}{1-(x+iy)} \right)=\frac{2}{5} <\frac{1}{2}$.
Hint:
Using your computations:
$$ \Re\left(\frac{1}{1-z}\right)=\frac{1-x}{(1-x)^2+y^2}=\frac{1-x}{x^2+y^2-2x+1}$$
but $x^2+y^2<1$ so:
$$\frac{1-x}{x^2+y^2-2x+1}>\frac{1-x}{1-2x+1}=\frac{1}{2}$$
