= Solution
Put $F(s)=-\zeta'(s)/\zeta(s)$. The <Euler product> gives $F(s)=\sum_{n\geq1}\Lambda(n)n^{-s}$ for $\Re s>1$. The elementary nonnegative polynomial
$$
3+4\cos u+\cos2u=2(1+\cos u)^2
$$
therefore gives the <three-four-one zero-free-region argument>:
$$
0\leq3F(\sigma)+4\Re F(\sigma+it)+\Re F(\sigma+2it)
\qquad(\sigma>1).
$$
The coefficients $\Lambda(n)$ are nonnegative; applying the identity with $u=t\log n$ justifies the inequality term by term.
Take real parts in the <global partial-fraction expansion of the zeta logarithmic derivative>. For $1<\sigma\leq2$, the <digamma function> bounds yield
$$
\Re F(\sigma+it)
\leq\Re\frac1{\sigma+it-1}
-\sum_\rho\frac{\sigma-\beta}{(\sigma-\beta)^2+(t-\gamma)^2}
+C\log(|t|+2).
$$
All terms in the zero sum are nonnegative, because the <Euler product> and completed <functional equation> locate the nontrivial zeros in the closed critical strip. The constant terms are bounded: in particular $\sum_\rho\Re(1/\rho)=\sum_\rho\beta/|\rho|^2$ converges. Near the real <pole>, $F(\sigma)=1/(\sigma-1)+O(1)$.
First exclude zeros on the line one. If $1+it$, $t\ne0$, were a zero of multiplicity $m\geq1$, the selected zero term would give
$$
\Re F(\sigma+it)\leq-\frac m{\sigma-1}+O_t(1),
\qquad \Re F(\sigma+2it)\leq O_t(1).
$$
Inserted into the positivity inequality, this yields $0\leq(3-4m)/(\sigma-1)+O_t(1)$, impossible as $\sigma\downarrow1$. At $t=0$, zeta has a <pole>, with a punctured neighbourhood containing no zeros. Thus the line one has no zeros.
Now let $\rho=\beta+i\gamma$ with $|\gamma|\geq1$ and put $L=\log(|\gamma|+2)$. Select this zero in the formula at $t=\gamma$ and discard every other nonnegative zero kernel. The <pole> terms at heights $\gamma,2\gamma$ are bounded. Consequently
$$
0\leq\frac3{\sigma-1}-\frac4{\sigma-\beta}+C_1L.
$$
Choose a fixed positive $a$ small enough that $C_1a\leq1/2$ and $1+a/L\leq2$, and set $\sigma=1+a/L$, $h=1-\beta$. Rearranging gives
$$
\frac4{a/L+h}\leq L\left(\frac3a+C_1\right),\qquad
h\geq\frac{a(1-C_1a)}{(3+C_1a)L}\geq\frac a{7L}.
$$
This is the required logarithmic gap at large height.
For $|\gamma|\leq1$, zeros with $\beta\leq1/2$ already have a fixed gap. The remaining compact region contains finitely many zeros, none on the line one; the <pole> at one cannot be an accumulation point of zeros. Hence those zeros also have a fixed positive distance from that line. Reduce the absolute constant to cover this bounded-height region and to make the preceding weak inequality strict. The <zero-free region of the Riemann zeta function> is therefore
$$
\boxed{\beta<1-\frac{c}{\log(|\gamma|+2)}
\quad\text{for every nontrivial zero, with an absolute }c>0.}
$$
This proof has established boundary nonvanishing as well as the quantitative gap, rather than assuming the former implicitly.
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