I'm trying to follow the proof in Wikipedia that the PNT is equivalent to the assertion $\psi(x)\sim x$, by proving that $\psi(x)\sim\pi(x)\log x$, which it claims is a very simple proof. One direction of inequality is an actual bound, $\psi(x)\le\pi(x)\log x$, but the other inequality has a fuzz factor:
$$\psi(x) \ge \sum_{x^{1-\epsilon}\le p\le x} \log p\ge\sum_{x^{1-\epsilon}\le p\le x}(1-\epsilon)\log x=(1-\epsilon)(\pi(x)+O(x^{1-\epsilon}))\log x.$$
But this doesn't actually complete the proof, because we want $\psi(x)\ge(1-\epsilon)\pi(x)\log x$ without the fuzz factor. If we take large enough $x$ and use $\epsilon/2$ in the above equation we get
$$\psi(x)\ge(1-\epsilon/2)\pi(x)\log x+Ax^{1-\epsilon/2}\log x,$$ so it is sufficient to prove that $Ax^{1-\epsilon/2}\le\frac{\epsilon}2\pi(x)$ for sufficiently large $x$, i.e. $x^{1-\epsilon/2}\in o(\pi(x)),$ and although I am sure there is a proof of this, it's not so simple that the proof can be completely omitted, at least as far as I can see. Is there an easy proof to be found here? The only one I am seeing is Chebyshev's weak version of the PNT, $\frac x{\log x}\in O(\pi(x))$, which takes some significant work to prove.
You have: $$\psi(x)\geq (1-\varepsilon)\sum_{x^{1-\varepsilon}<p<x}\log x \geq (1-\varepsilon)(\pi(x)-\pi(x^{1-\varepsilon}))\log x$$ but, by the Chebyshev bound: $$\frac{\pi(x^{1-\varepsilon})}{\pi(x)}\leq\frac{1.11\frac{x^{1-\varepsilon}}{(1-\varepsilon)\log x}}{0.92\frac{x}{\log x}}\leq \frac{1.21}{1-\varepsilon}\, x^{-\varepsilon}\tag{1}$$ is enough to prove that: $$\psi(x)\geq (1-\tilde{\varepsilon})\,\pi(x)\log x.$$ To replace $(1)$ with something easier to prove than the Chebyshev bound, you can use $\pi(x^{1-\varepsilon})\leq x^{1-\varepsilon}$ and any suitable lower bound for $\pi(x)$.