<s>[INST] 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.
 [/INST] $$\Re \Big(\frac{1}{1-z}\Big)={1\over 2}\Big(\frac{1}{1-z}+\frac{1}{1-\overline{z}}\Big)$$
$$={1\over 2}\frac{1-\overline{z}+1-z}{1-z-\overline{z}+z\overline{z}}$$
$$>{1\over 2}\frac{1-\overline{z}+1-z}{1-z-\overline{z}+1}$$
$$={1\over 2}$$
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