 Solution

Complete multiplicativity and $f(p)=\pm 1$ give, for $\mathrm{Re}s>1$,The product is nonzero there; its absolutely convergent logarithm has terms $f(p{)}^m{p}^{-ms}/m$. ThusEach numerator is one of $3\pm 4\cos \theta +\cos 2\theta$, so it is nonnegative. Exponentiation gives the required lower bound one.

If $F(1+it)=0$ with $t\ne 0$, its assumed holomorphy makes $|F(\sigma +it){|}^4=O((\sigma -1{)}^4)$, or smaller for a higher-order zero. The zeta factor at real $\sigma$ grows at most as $(\sigma -1{)}^{-3}$, while the zeta factor near $1+2it$ stays bounded. The product tends to zero, a contradiction. Hence