This is elementary, the negation of your statement is $\psi(x)-x=o(x^{1/2})$ which implies that $$F(s)=s\int_1^\infty (\psi(x)-x) x^{-s-1}dx= \frac{-\zeta'(s)}{\zeta(s)}-\frac1{s-1}$$ is analytic for $\Re(s) >1/2$ and as $\Re(s)\to 1/2$ $$F(s) = s\int_1^\infty o(x^{-s-3/2})dx= o(\frac{s}{\Re(s)-1/2})$$

Contradicting that $\frac{\zeta'(s)}{\zeta(s)}$ has some poles on $\Re(s)=1/2$ ie. contradicting that the $\Bbb{R\to R}$ function $\pi^{-(1/2+it)/2}\Gamma((1/2+it)/2)\zeta(1/2+it)$ has a sign change around $t=14.1$