We know that the Riemann zeta function $\zeta$ has no zeros in the region $\{\beta+it:\beta>1-\frac{c}{\log(2+|t|)}\}$, where $c>0$ is an absolute constant. This is known as the classical zero-free region of zeta function.
Now I need to bound the log derivative of zeta function in the region. Namely, show that there is an absolute constant $c_1>0$ such that we have the bound $$\frac{\zeta'}{\zeta}(\sigma+it)=O(\log (2+|t|))$$ whenever $|t|\gg 1$ and $\sigma\geq 1-\frac{c_1}{\log(2+|t|)}$.
From the explicit formula, we have $$-\frac{\zeta'}{\zeta}(s)=-\sum_{\rho:|\rho-s|\leq\varepsilon/2}\frac{1}{s-\rho}+\frac{1}{s-1}+O_{C,\varepsilon}(\log (2+|t|))$$ whenever $s=\sigma+it$ and $\varepsilon\leq\sigma\leq C$. If we set $c_1=c/2$ and use the fact that zeta function has at most $O_\varepsilon(\log(2+|t|))$ zeros in the region $\{\sigma+it:\varepsilon\leq \sigma\leq 1:|t-t_0|\leq 1\}$ to remove the contribution of zeroes further than $c_1/\log(2+|t|)$ from $s$, then we have $$\frac{\zeta'(\sigma+it)}{\zeta(\sigma+it)}=O(\log^2(2+|t|))$$ whenever $|t|\gg 1$ and $\sigma\geq 1-\frac{c_1}{\log(2+|t|)}$. But how to reduce the power to $1$?