I'm currently working on the following problem from Analytic Number Theory.
Assume that $\psi(x)-x=\mathcal{O}(x^a)$, for some $1/2<a<1$, where $\psi$ is the Chebyshev function.
I would like to show that this implies that $\zeta(s)$ has no zeroes with $\sigma>a$, where $s=\sigma+it$.
I'm not sure how to proceed with the proof of this claim. Any hint/help will be very useful. Thank you.