= Solution
<Absolute convergence> and the <Fundamental theorem of arithmetic> give the <Euler product>
$$
\boxed{\zeta(s)=\prod_{p\ \mathrm{prime}}(1-p^{-s})^{-1},\qquad\Re s>1.}
$$
For a finite set of primes, expand the geometric factors: their product sums $n^{-s}$ over integers whose prime factors lie in that set. Let the finite sets increase through all primes. <Absolute convergence> permits passage to the limit and recovers the full <Dirichlet series>. Moreover $\sum_{p,m\ge1}|p^{-ms}|/m<\infty$, so the logarithm converges and the product has no zeros there.
The same <absolutely convergent> logarithm, and $3+4\cos\theta+\cos2\theta=2(1+\cos\theta)^2\ge0$, give
$$
\zeta(\sigma)^3|\zeta(\sigma+it)|^4|\zeta(\sigma+2it)|\ge1\quad(\sigma>1).
$$
This is the product version of the <three-four-one zero-free-region argument>. It also proves there are no zeros on $\Re s=1$: if $\zeta(1+it_0)=0$ for $t_0\ne0$, its factor has order at least four as $\sigma\downarrow1$, while the real <pole> contributes only order minus three and the $2t_0$ factor remains bounded. The displayed left side would tend to zero, a contradiction. At $t_0=0$ there is a <pole>, not a zero.
For large $|t|$, put $L=\log(|t|+2)$. The <Hardy-Littlewood approximation to the Riemann zeta function> at $x\asymp|t|$ gives $|\zeta(\sigma+it)|\ll L$ for $1-2/L\le\sigma\le3$; its finite sum is bounded by $\sum_{n\le x}n^{-1+2/L}\ll L$ and the integral term is bounded. The <Cauchy estimate for derivatives> on circles of radius comparable to $1/L$ consequently gives $|\zeta'(\sigma+it)|\ll L^2$ for $1-L^{-9}\le\sigma\le2$.
Take $\sigma_0=1+aL^{-9}$ with a small fixed $a>0$. The product inequality, $\zeta(\sigma_0)\ll L^9/a$, and $|\zeta(\sigma_0+2it)|\ll L$ imply
$$
|\zeta(\sigma_0+it)|\ge c_0a^{3/4}L^{-7}.
$$
If $1-cL^{-9}\le\sigma\le\sigma_0$, integration of the <derivative> along the horizontal segment changes this value by at most $C(a+c)L^{-7}$. Choose $a$ sufficiently small that $Ca<c_0a^{3/4}/4$, then $c\le a$ sufficiently small. The lower bound remains a positive multiple of $L^{-7}$. For $\sigma_0\le\sigma\le2$, the same product inequality, $\zeta(\sigma)\le\zeta(\sigma_0)$, and the near-one upper bound give that lower bound directly. For $\sigma\ge2$, the reciprocal <Euler product> gives $|1/\zeta(s)|\le\zeta(2)$.
Finally the no-zero result on $\Re s=1$, <compactness> at bounded heights and the regular reciprocal at the <pole> allow a further fixed reduction of $c$ to include bounded $t$. We have proved the <weak logarithmic zero-free region for the Riemann zeta function>
$$
\boxed{\left|\frac1{\zeta(\sigma+it)}\right|\ll\log^7(|t|+2),\qquad\sigma\ge1-\frac c{\log^9(|t|+2)}.}
$$
The reciprocal at $s=1$ is its holomorphic extension, equal to zero.
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