= Solution
Write $F(s)=-\zeta'(s)/\zeta(s)$ and $\psi_\Gamma=\Gamma'/\Gamma$. The <global partial-fraction expansion of the zeta logarithmic derivative> is
$$
F(s)=\frac1{s-1}-\frac12\log\pi+\frac12\psi_\Gamma(1+s/2)-B-\sum_\rho\left(\frac1{s-\rho}+\frac1\rho\right),\qquad B=\log2+\frac12\log\pi-1-\frac\gamma2.
$$
Here $\gamma$ is the <Euler--Mascheroni constant>, and the sum includes multiplicities of the <Nontrivial zeros of the Riemann zeta function>. The paired terms are locally absolutely convergent away from their <poles>, by the <Riemann–von Mangoldt formula>. This identity follows by differentiating the genus-one <Hadamard factorization> of the <Riemann xi function> and using the <Gamma function recurrence>.
We first prove the classical <zero-free region of the Riemann zeta function>. The <Euler-product nonvanishing of the Riemann zeta function> and its <functional equation> place every nontrivial zero in the closed strip $0\le\beta\le1$; no assertion about zeros on its boundary is needed yet. The <real logarithmic derivative of Riemann xi> and the <digamma function> asymptotic imply, for $1<\sigma\le2$ and $|t|\ge1$,
$$
\Re F(\sigma+it)\le C\log(|t|+2)-\sum_\rho\frac{\sigma-\beta}{(\sigma-\beta)^2+(t-\gamma_\rho)^2}.
$$
Every summand being subtracted is nonnegative. If $\beta+it$ is a zero, retaining its summand gives $\Re F(\sigma+it)\le C\log(|t|+2)-1/(\sigma-\beta)$; at height $2t$ the upper bound $C\log(|t|+2)$ suffices. At real $\sigma$ the simple <pole> of $\zeta$ gives $F(\sigma)=1/(\sigma-1)+O(1)$.
For $\sigma>1$, the <Dirichlet series> $F(s)=\sum_n\Lambda(n)n^{-s}$ has nonnegative coefficients. The identity $3+4\cos u+\cos2u=2(1+\cos u)^2\ge0$ therefore gives the <three-four-one zero-free-region argument>
$$
0\le3F(\sigma)+4\Re F(\sigma+it)+\Re F(\sigma+2it).
$$
With $L=\log(|t|+2)$, a zero at $\beta+it$ would consequently satisfy
$$
\frac4{\sigma-\beta}\le\frac3{\sigma-1}+CL.
$$
Choose $\sigma=1+a/L$, with an absolute $a>0$ sufficiently small, and put $c=a/8$. If $\beta\ge1-c/L$, this inequality would imply $4/(a+c)\le3/a+C$. But $4/(a+a/8)-3/a=5/(9a)>C$ for small enough $a$, a contradiction.
For any fixed $t\ne0$, the same positivity inequality as $\sigma\downarrow1$ excludes a zero at $1+it$: its negative contribution has coefficient at least four, whereas the real <pole> contributes only three. Any zero at $1+2it$ only strengthens the contradiction. The <pole> at $s=1$ has a zero-free punctured neighbourhood. By compactness, the remaining bounded-height portion of the line has a zero-free neighbourhood as well. Shrink $c$ to combine these facts with the large-height bound. Hence
$$
\boxed{\zeta(\beta+i\gamma)\ne0\quad\text{if}\quad\beta\ge1-\frac c{\log(|\gamma|+2)},}
$$
with the understanding that $s=1$ is a <pole>.
The <Riemann–von Mangoldt explicit formula> for the <Second Chebyshev function>, with half weight at a prime-power endpoint, is
$$
\psi_0(x)=x-\lim_{T\to\infty}\sum_{|\Im\rho|\le T}\frac{x^\rho}{\rho}-\log(2\pi)-\frac12\log(1-x^{-2}),\qquad x>1.
$$
The zero sum uses symmetric height truncation. To estimate it, take a half-integer $x$ and use the <truncated explicit formula for the second Chebyshev function>
$$
\psi(x)=x-\sum_{|\Im\rho|\le T}\frac{x^\rho}{\rho}+O\left(\frac{x\log^2(xT)}T+\log x\right),\qquad 2\le T\le x.
$$
The zero-free region gives $|x^\rho|\le x\exp(-c\log x/\log(T+2))$ for these zeros. The <Riemann–von Mangoldt formula> and <partial summation> give $\sum_{|\Im\rho|\le T}|\rho|^{-1}\ll\log^2(T+2)$. Choose $T=\exp(\sqrt{\log x})$. Both the zero sum and the truncation error are then $O(xe^{-c'\sqrt{\log x}})$ after reducing the positive constant to absorb logarithmic factors. Replacing an arbitrary $X$ by $\lfloor X\rfloor+1/2$ preserves the <Von Mangoldt function> sum and changes the main term by at most one. This proves the requested form of the <Prime number theorem>:
$$
\boxed{\sum_{n\le X}\Lambda(n)=X+O\left(Xe^{-c'\sqrt{\log X}}\right).}
$$
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