For $\sigma>1$, take logarithms of Euler products. With $z=f(p)p^{-it}$, the inequality
$$
3+4\Re(z^k)+\Re(z^{2k})\geq0
\qquad(|z|\leq1)
$$
at every prime power gives the <three-four-one inequality for Euler products>
$$
\zeta(\sigma)^3|D_f(\sigma+it)|^4|D_{f^2}(\sigma+2it)|\geq1.
$$

Suppose $D_f(1+it)=0$ with multiplicity $m\geq1$. The assumed analytic continuation gives
$$
|D_f(\sigma+it)|\ll(\sigma-1)^m,
$$
while $D_{f^2}(\sigma+2it)$ remains bounded and $\zeta(\sigma)^3\asymp(\sigma-1)^{-3}$ as $\sigma\downarrow1$. The left side of the inequality would then be
$$
O((\sigma-1)^{4m-3})\longrightarrow0,
$$
contradicting its lower bound one. Hence $D_f$ has no zero on $\Re s=1$. Absolute convergence of its Euler product already excludes zeros for $\Re s>1$, so $D_f(s)\ne0$ throughout $\Re s\geq1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 150 / 3 / d / Solution

For every complex number $z$ with $|z|\leq1$,
$$
3+4\Re z+\Re(z^2)\geq0.
$$
Applying this to each prime-power term in logarithms of Euler products proves that
$$
\zeta(\sigma)^3|D_f(\sigma+it)|^4|D_{f^2}(\sigma+2it)|\geq1
$$
for $\sigma>1$ and every completely multiplicative $f$ bounded by one.


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