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If $f$ is holomorphic near the closed radius-$R$ disc, $f(z_0)\ne 0$, and $|f|\le M$, the displayed estimate holds away from zeros for $|z-z_0|\le R/3$, counting zeros with multiplicity. Factoring local zeros isolates their poles; the remaining logarithm is controlled by disc estimates. If all zeros lie to the left of a point $z$ in real part, their real contributions are nonnegative. This is the disc estimate used by the Landau zero-free-region theorem; a version with general radius ratios is Lemma 24.17 in Montgomery and Vaughan.

Here is a quantitative local form of the Landau zero-free-region theorem. Let $t\ge 2$, $0<\eta \le 1/4$, $M\ge 2$, and suppose $|\zeta (z)|\le M$ on the two closed discs of radius $\eta$ centred at $1+\eta /4+it$ and $1+\eta /4+2it$. Then every zero $\beta +it$ satisfiesfor an absolute positive $c$. In particular, if upper bounds of this form hold locally for every large height, they give a zero-free region of this width. The discs are away from the pole, and the Euler product excludes zeros to the right of one.

We state precisely the permitted local logarithmic-derivative lemma. If $f$ is holomorphic on a neighborhood of $|z-z_0|\le R$, $f(z_0)\ne 0$, and $|f|\le M$, then for $|z-z_0|\le R/3$ away from zeros,The zeros are counted with multiplicity. This standard disc estimate, which may be assumed here, follows by factoring nearby zeros and applying a Cauchy estimate for derivatives to the remaining logarithm. When every zero has real part at most one and $\Re z>1$, the zero terms have nonnegative real parts, giving the required lower bound. The estimate with fixed radius ratios is also recorded as Lemma 24.17 in Montgomery and Vaughan's general treatment.

The reciprocal Euler product gives $|\zeta (1+\eta /4+ijt)|\ge 1/\zeta (1+\eta /4)\gg \eta$ for $j=1,2$. Put $B=1+\log (M/\eta )$. The lemma's error on both discs is therefore $O(B/\eta )$. Let $d=1-\beta$. If $d\ge \eta /24$, the desired conclusion already holds after reducing $c$. Otherwise $\beta +it$ is among the local zeros and, for $1<\sigma \le 1+\eta /4$,All other zero terms may be discarded because their real parts are nonnegative. The simple pole at one gives $-{\zeta }^{\prime }(\sigma )/\zeta (\sigma )=(\sigma -1{)}^{-1}+O(1)$. Insert these inequalities into part (a):There is no zero on the line one by the argument in Question 2(a), so $d>0$. Choose $\sigma =1+6d$, which lies in the indicated range. The left side is $(4/7-3/6)/d=1/(14d)$. Hence $d\gg \eta /B$, proving the theorem. The logarithm of the upper bound, rather than the upper bound itself, is what enters the zero-free width.

Put $L=\log t$ and $\ell =\log L$, for sufficiently large $t$. Apply the Landau zero-free-region theorem withwhere $a>0$ is fixed and small enough that $\eta \le 1/4$. On its two discs, the real part is at least $1-3\eta /4$ and the imaginary part is comparable to $t$. The given Richert bound for the Riemann zeta function therefore gives, on the part left of one,On the part right of one, the separately given $O({\log }^{2/3}t)$ bound gives the same conclusion. We may thus choose $M=L^A$ for one fixed sufficiently large $A$. Also $\log (1/\eta )=O(\ell )$, so the logarithmic term in the Landau zero-free-region theorem is $O(\ell )$. Its conclusion isfor large $t$ and a sufficiently small positive $c$. Complex conjugation supplies negative heights. This is the Vinogradov-Korobov zero-free region. Only the stated Richert upper bounds, the Euler product, the pole at one and the proved Landau zero-free-region theorem were used; no prior zero-free-region theorem was assumed.

For large positive $t$ and $0<\sigma \le 1$, a fixed sufficiently large $C$ gives the displayed upper bound; to the right of one use $|\zeta (\sigma +it)|\ll {\log }^{2/3}t$. Its nonlinear dependence on $1-\sigma$ permits a wider zero-free region of the Riemann zeta function through the Landau zero-free-region theorem. Its proof uses estimates for exponential sums; its use as an upper-bound input is separate from a proof of a zero-free region.

For sufficiently large $|t|$, the Riemann zeta function has no zeros in the displayed region, for a fixed positive $c$. To derive it from the Richert bound for the Riemann zeta function, take $\eta \asymp (\log \log t/\log t{)}^{2/3}$. The logarithm of the maximum on the two discs in the Landau zero-free-region theorem is $O(\log \log t)$, as is $\log (1/\eta )$. Dividing $\eta$ by this logarithmic factor gives the displayed width.