In section 2.5 of the book "The Riemann zeta function" by H.M. Edwards, is stated:
Since all but a finite number of roots $\rho$ satisfy the inequality $$\frac{1}{|\rho(1-\rho)|}=\frac{1}{\left|\left(\rho-\frac{1}{2}\right)^2-\frac{1}{4}\right|}\lt \frac{1}{\left|\rho-\frac{1}{2}\right|^2}$$
I cannot see how this is true. I can easily prove it if I replace $\displaystyle \frac{1}{\left|\rho-\frac{1}{2}\right|^2}$ with $\displaystyle \frac{2}{\left|\rho-\frac{1}{2}\right|^2}$, because there can only be finitely many zeros inside the circle $|z|\leq 1/2$. But it seems to me that the boundary of the region determined by $$\frac{1}{\left|\left(\rho-\frac{1}{2}\right)^2-\frac{1}{4}\right|}= \frac{1}{\left|\rho-\frac{1}{2}\right|^2}$$ is not bounded, so I do not understand the statement. Thanks for any clarification.
Shortly after I posted my question I realized what is happening: zeros can only be in the strip $0\lt \mbox{Re}\rho \lt 1$, and so the only regions of the complex plane where the inequality can fail are bounded sets, and an analytic function can only have finitely many zeros in a bounded set.