$| \Re(\tau)| \leq 1/2$ for $\tau$ in fundamental region.

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I've been working through Alfors section on elliptic functions and I'm stuck on an inequality in one of the proofs regarding the fundamental region (thereom 2 of 2.3 The canonical basis). We have the following setup: $w_1, w_2 \in \mathbb{C}$ and their ratio is defined $\tau = \frac{w_2}{w_1}$. We have chosen them so that $$|w_1| \leq |w_2|$$ $$\Im (\tau) \neq 0$$$$ |w_2| \leq |w_1+w_2|$$$$ |w_2| \leq |w_1-w_2|$$ Alfors then says that inequalities these are equivalent to $|\tau| \geq 1$ (which is obvious) and $|\Re(\tau)| \leq 1/2,$ which is the inequality I am stuck on.

I have tried writing the $w_i$ as $a+bi$ and $c+di$ and using that $\Re(\tau)$ is of the form $\dfrac{ac+bd}{c^2+d^2}$, but I'm not sure this is the right approach.

Thanks in advance!

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Write down $\tau=a+ib$, with $a,b\in \mathbb{R}$. We know that $$|\tau|^2\leq\min(|1+\tau|^2,|1-\tau|^2)$$

But $|1+\tau|^2=(a+1)^2+b^2$ and $|1-\tau|^2=(a-1)^2+b^2$. So actually, we have $$a^2\leq \min((a+1)^2,(a-1)^2)$$ Which is equivalent to $$0\leq \min(2a+1,-2a+1)$$ This eventually exactly means $|a|\leq \frac{1}{2}$.

Now, to be complete, I let you try and prove the converse!