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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.

Set $\epsilon =1-\sigma \in [0,1)$ and $L=\log (t+2)$. By the Hardy-Littlewood approximation to the Riemann zeta function at $x=t$, it is enough to bound ${\sum }_{n\le t}{n}^{-\sigma -it}$: the integral term has size $O(t^{-\sigma })$ because $|s-1|\ge t$.

On a dyadic interval $(N,N^{\prime }]\subset (N,2N]$ with $N^{\prime }\le t$, the assumed estimate holds for every initial subinterval. Abel summation with $n^{-\sigma }$ therefore givesIndeed the weighted endpoint and integral of the term proportional to the subinterval length are $O(N^{1-\sigma })$, and those of the constant $N^{4/5}$ term are $O(N^{4/5-\sigma })$. The constants can be uniform in $0<\sigma \le 1$.

For the first term, write $u=j\log 2$ and complete the square:The sum of a shifted Gaussian function on a fixed-spaced lattice is $O(\sqrt{L})$, uniformly in the shift. Thus these dyadic contributions are $O(\sqrt{L}{e}^{{\epsilon }^{2}L/(4c)})$. This is the Gaussian dyadic summation bound.

For the second term, if $\epsilon \le 1/10$ its dyadic sum is bounded. If $\epsilon >1/10$, a crude bound is $O(L\exp (\max (\epsilon -1/5,0)L))$. The positive exponent obeys $\epsilon -1/5\le (5/4){\epsilon }^2$, and $L$ can be absorbed into $\sqrt{L}{e}^{C{\epsilon }^{2}L}$ uniformly on $\epsilon >1/10$ by increasing the fixed constant $C$. The finitely many initial terms cause no problem. We conclude, with one fixed sufficiently large $C$,In particular the endpoint $\sigma =1$ gives $O(\sqrt{\log t})$ under the assumed exponential-sum hypothesis. This conditional conclusion uses that hypothesis, not an unconditional improvement of the stated Richert bound for the Riemann zeta function.

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.