= Solution
The <double-angle formula> gives
$$
3\pm4\cos\theta+\cos2\theta=2(1\pm\cos\theta)^2\ge0.
$$
For $\sigma>1$, the <logarithm> of the absolutely convergent <Euler product> gives
$$
\begin{aligned}
&3\log\zeta(\sigma)+4\log|\zeta(\sigma+it)|+\log|\zeta(\sigma+2it)|\\
&=\sum_p\sum_{m\ge1}\frac{3+4\cos(mt\log p)+\cos(2mt\log p)}{mp^{m\sigma}}\ge0.
\end{aligned}
$$
Exponentiation proves
$$
\boxed{\zeta(\sigma)^3|\zeta(\sigma+it)|^4|\zeta(\sigma+2it)|\ge1.}
$$
Suppose $\zeta(1+it)=0$ for $t\ne0$, of order $r\ge1$. As $\sigma\downarrow1$, its middle factor is $O((\sigma-1)^{4r})$, the first is $O((\sigma-1)^{-3})$ because the <pole> at one is simple, and the final factor is bounded because $1+2it\ne1$. The product is $O((\sigma-1)^{4r-3})\to0$, contradicting the lower bound one. Therefore
$$
\boxed{\zeta(1+it)\ne0\qquad(t\ne0).}
$$
This is the <three-four-one zero-free-region argument> specialized to the line with real part one.
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