Chebyshev bounds for the prime-counting function

By [L2], choose positive constants $a,b$ and $y_0\ge 2$ such that $at\le \theta (t)\le bt$ for every $t\ge y_0$. Enlarging $b$ if needed to absorb the finite range $2\le t\le y_0$, we may assume $\theta (t)\le bt(t\ge 2)\mathrm{.}$

For $x\ge y_0$, the integral term in [L1] is nonnegative, so $\pi (x)\ge \frac{\theta (x)}{\log x}\ge a\frac{x}{\log x}\mathrm{.}$

Assume now that $x\ge \max \{y_0^2,e^2\}$. Using [L1] and step 1.1, $\pi (x)\le b\frac{x}{\log x}+b{\int }_2^x\frac{dt}{{\log }^{2}t}\mathrm{.}$ Split the integral at $\sqrt{x}$. On $[2,\sqrt{x}]$ one has $\log t\ge \log 2$, so ${\int }_2^{\sqrt{x}}\frac{dt}{{\log }^{2}t}\le \frac{\sqrt{x}}{{\log }^{2}2}<3\sqrt{x}\mathrm{.}$ On $[\sqrt{x},x]$ one has $\log t\ge \frac{1}{2}\log x$, so ${\int }_{\sqrt{x}}^x\frac{dt}{{\log }^{2}t}\le \frac{4x}{{\log }^{2}x}\mathrm{.}$ Hence $\pi (x)\le b\frac{x}{\log x}+3b\sqrt{x}+\frac{4bx}{{\log }^{2}x}\mathrm{.}$

Since $\sqrt{x}=o(x\mathrm{/}\log x)$ and $x\mathrm{/}{\log }^{2}x\le x\mathrm{/}\log x$ for $x\ge e$, step 2.2 implies $\pi (x)\le c_2\frac{x}{\log x}$ for some positive constant $c_2$ and all sufficiently large $x$. Taking $c_1=a$ and enlarging $x_0$ if necessary to satisfy both steps 2.1 and 2.2 proves the theorem.