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It is elementary, take $K=1/30 < 1/|\rho_0|$ where $\rho_0\approx 1/2+i14.13$ is the first zero, if $\psi(x)-x\le Kx^{1/2}$ for $x$ large enough then $x-\psi(x)+K x^{1/2}+C\ge 0$ for all $x$, where $C$ is a suitable real constant. Let $$F(s)=\int_1^\infty (x-\psi(x)+Kx^{1/2}+C) x^{-s-1}dx= \frac1{s-1}+\frac{\zeta'(s)}{s\zeta(s)}+\frac{K}{s-1/2}+\frac{C}{s}$$ By the non-negativity of the integrand it has a singularity at its abscissa of convergence $\sigma$. But the RHS is analytic on $(1/2,\infty)$ thus $\sigma=1/2$. And by the non-negativity again as $\Re(s) \to 1/2$ $$|F(s)|\le F(\Re(s))\sim \frac{K}{\Re(s)-1/2}$$ Contradicting that $$F(s)\sim\frac{\zeta'(s)}{s\zeta(s)}\sim\frac{1/\rho_0}{s-\rho_0} \text{ as } s\to \rho_0$$

It is elementary, take $K=1/30 < 1/|\rho_0|$ where $\rho_0\approx 1/2+i14.13$ is the first zero, if $\psi(x)-x\le Kx^{1/2}$ for $x$ large enough, then $x-\psi(x)+K x^{1/2}+C\ge 0$ for all $x$, where $C$ is a suitable real constant. Let $$F(s):=\int_1^\infty (x-\psi(x)+Kx^{1/2}+C) x^{-s-1}dx= \frac1{s-1}+\frac{\zeta'(s)}{s\zeta(s)}+\frac{K}{s-1/2}+\frac{C}{s}.$$ By the non-negativity of the integrand, it has a singularity at its abscissa of convergence $\sigma$. But the RHS is analytic on $(1/2,\infty)$, thus $\sigma=1/2$. And by the non-negativity again we have, as $\Re(s) \to 1/2$, $$|F(s)|\le F(\Re(s))\sim \frac{K}{\Re(s)-1/2}.$$ This contradicts that $$F(s)\sim\frac{\zeta'(s)}{s\zeta(s)}\sim\frac{1/\rho_0}{s-\rho_0}\qquad\text{as $s\to \rho_0$}.$$

It is elementary, take $K=1/30 < 1/|\rho_0|$ where $\rho_0\approx 1/2+i14.13$ is the first zero, if $\psi(x)-x\le Kx^{1/2}$ for $x$ large enough then $x-\psi(x)-K x^{1/2}+C\ge 0$. Let $$F(s)=\int_1^\infty (x-\psi(x)+Kx^{1/2}+C) x^{-s-1}dx= \frac1{s-1}+\frac{\zeta'(s)}{s\zeta(s)}+\frac{K}{s-1/2}+\frac{C}{s}$$ By the non-negativity of the integrand it has a singularity at its abscissa of convergence $\sigma$. But the RHS is analytic on $(1/2,\infty)$ thus $\sigma=1/2$. And by the non-negativity again as $\Re(s) \to 1/2$ $$|F(s)|\le F(\Re(s))\sim \frac{K}{\Re(s)-1/2}$$ Contradicting that $$F(s)\sim\frac{\zeta'(s)}{s\zeta(s)}\sim\frac{1/\rho_0}{s-\rho_0} \text{ as } s\to \rho_0$$

It is elementary, take $K=1/30 < 1/|\rho_0|$ where $\rho_0\approx 1/2+i14.13$ is the first zero, if $\psi(x)-x\le Kx^{1/2}$ for $x$ large enough then $x-\psi(x)+K x^{1/2}+C\ge 0$ for all $x$, where $C$ is a suitable real constant. Let $$F(s)=\int_1^\infty (x-\psi(x)+Kx^{1/2}+C) x^{-s-1}dx= \frac1{s-1}+\frac{\zeta'(s)}{s\zeta(s)}+\frac{K}{s-1/2}+\frac{C}{s}$$ By the non-negativity of the integrand it has a singularity at its abscissa of convergence $\sigma$. But the RHS is analytic on $(1/2,\infty)$ thus $\sigma=1/2$. And by the non-negativity again as $\Re(s) \to 1/2$ $$|F(s)|\le F(\Re(s))\sim \frac{K}{\Re(s)-1/2}$$ Contradicting that $$F(s)\sim\frac{\zeta'(s)}{s\zeta(s)}\sim\frac{1/\rho_0}{s-\rho_0} \text{ as } s\to \rho_0$$

It is elementary, take $K=1/30 < 1/|\rho_0|$ where $\rho_0\approx 1/2+i14.13$ is the first zero, if $\psi(x)-x\le Kx^{1/2}$ for $x$ large enough then $x-\psi(x)-K x^{1/2}+C\ge 0$. Let $$F(s)=\int_1^\infty (x-\psi(x)-Kx^{1/2}+C) x^{-s-1}dx= \frac1{s-1}+\frac{\zeta'(s)}{s\zeta(s)}-\frac{K}{s-1/2}+\frac{C}{s}$$ By the non-negativity of the integrand it has a singularity at its abscissa of convergence $\sigma$. But the RHS is analytic on $(1/2,\infty)$ thus $\sigma=1/2$. And by the non-negativity again as $\Re(s) \to 1/2$ $$|F(s)|\le F(\Re(s))\sim \frac{K}{\Re(s)-1/2}$$ Contradicting that $$F(s)\sim\frac{\zeta'(s)}{s\zeta(s)}\sim\frac{1/\rho_0}{s-\rho_0} \text{ as } s\to \rho_0$$

It is elementary, take $K=1/30 < 1/|\rho_0|$ where $\rho_0\approx 1/2+i14.13$ is the first zero, if $\psi(x)-x\le Kx^{1/2}$ for $x$ large enough then $x-\psi(x)-K x^{1/2}+C\ge 0$. Let $$F(s)=\int_1^\infty (x-\psi(x)+Kx^{1/2}+C) x^{-s-1}dx= \frac1{s-1}+\frac{\zeta'(s)}{s\zeta(s)}+\frac{K}{s-1/2}+\frac{C}{s}$$ By the non-negativity of the integrand it has a singularity at its abscissa of convergence $\sigma$. But the RHS is analytic on $(1/2,\infty)$ thus $\sigma=1/2$. And by the non-negativity again as $\Re(s) \to 1/2$ $$|F(s)|\le F(\Re(s))\sim \frac{K}{\Re(s)-1/2}$$ Contradicting that $$F(s)\sim\frac{\zeta'(s)}{s\zeta(s)}\sim\frac{1/\rho_0}{s-\rho_0} \text{ as } s\to \rho_0$$