Consider the following question: The Möbius function is defined for square-free integers $n = p_1,...,p_k$ as $\mu(n) = (-1)^k$ ($p_i $ are distinct primes) and $\mu(n) = 0$ otherwise. The summatory function of $\mu(n)$ is $M(x) :=\sum_{n\leq x} \mu(n)$. Show that if we are given $0 < \delta < 1$ so that $M(x) = O(x^{\delta})$ for all $x \gg 1$, then $\zeta(s) \neq 0$ for all $s$ in the half-plane $\operatorname{Re}(s) > \delta$.
I am unable to make any progress on this and would appreciate if someone will tell me which result/theorem should I use.
I have completely studied Apostol's Introduction to Analytic Number Theory.
Consider $$f(s)=\sum_{n=1}^\infty\frac{\mu(n)}{n^s}.$$ By assumption this defines a holomorphic function in $\operatorname{Re}s>\delta$.
What is the relation between $\zeta(s)$ and $f(s)$ for $\operatorname{Re}s>1$ (Möbius inversion!)?
Now apply the identity theorem to conclude that this relation also holds for $\operatorname{Re}s>\delta$ (and $s\ne1$) and deduce that $\zeta(s)\ne0$ in this region.