The proof is based on inequalities for the function \begin{equation}\label{20150923.1} V(x) = T(x) - T\left( \frac{x}{2} \right)- T\left( \frac{x}{3} \right)- T\left( \frac{x}{6} \right) \end{equation} where \begin{equation*} T(x) = \sum_{ 1 \le n \le x} \log n \end{equation*} If \( \psi(x) = \sum_{n \le x } \Lambda(n) \) with \( \Lambda \) the von Mangoldt function, then \begin{equation*} T(x) = \sum_{n=1}^{\infty} \psi\left(\frac{x}{n} \right) \end{equation*} and thus \begin{equation}\label{20150923.2} V(x) = \psi(x) +\psi\left( \frac{x}{5} \right)- 2 \psi\left( \frac{x}{6} \right)+ \psi\left( \frac{x}{7} \right) +\psi\left( \frac{x}{11} \right)- 2 \psi\left( \frac{x}{12} \right) + \cdots \end{equation} The coefficients of \( \psi\left( \frac{x}{n} \right) \) in this expression have period 6. In contrast to a similar equation in Chebyshev's MÃ©moire, the signs in \eqref{20150923.2} are not alternating, this makes it less straightforward to use \eqref{20150923.2} to obtain bounds on \( \psi(x) \).

For all real \( x \ge 1 \), one has \begin{equation}\label{20150923.3} x \log x - x - \log x +1 \le T(x) \le x \log x - x + \log x +1 \end{equation} From \eqref{20150923.1} and \eqref{20150923.3} one immediately deduces that for all real \( x \ge 6 \) \begin{equation}\label{20150923.4} V(x) \le B x +4 \log x - 2 -2 \log 6 \end{equation} and \begin{equation}\label{20150923.5} B x - 4 \log x - 2 + 2 \log 6 \le V(x) \end{equation} where \( B = \frac{1}{2} \log 2 + \frac{1}{3} \log 3 + \frac{1}{3} \log 3\).

__Calculation of upper bound on \( \psi(x) \)__

Because \( n \to \psi\left( \frac{x}{n} \right) \) is decreasing, it follows that \begin{equation*} \psi\left( \frac{x}{7} \right)+\psi\left( \frac{x}{11} \right)- 2 \psi\left( \frac{x}{12} \right) \ge 0 \end{equation*} and \begin{equation*} \psi\left( \frac{x}{13} \right)+\psi\left( \frac{x}{17} \right)- 2 \psi\left( \frac{x}{18} \right) \ge 0 \end{equation*} and so on. Therefore from \eqref{20150923.2} one obtains \begin{equation*} V(x) \ge \psi(x) +\psi\left( \frac{x}{5} \right)- 2 \psi\left( \frac{x}{6} \right) \end{equation*} Because \( \psi\left( \frac{x}{5} \right)- \psi\left( \frac{x}{6} \right) \ge 0 \), it follows also that \begin{equation}\label{20150923.6} V(x) \ge \psi(x) - \psi\left( \frac{x}{6} \right) \end{equation} One can obtain an upper bound on \( \psi(x) \) with a standard calculation: combining \eqref{20150923.4} and \eqref{20150923.6} and using \(-2 - 2 \log 6 < -5 \), one has for all real \( x \ge 6 \) that \begin{equation}\label{20150923.7} \psi(x) - \psi\left(\frac{x}{6} \right)\le B x + 4 \log x - 5 \end{equation} Take \( m \in \mathbb{N} \) such that \( 6 \le \frac{x}{6^m} < 6^2 \). Applying \eqref{20150923.7} several times gives \begin{align*} \psi\left(\frac{x}{6} \right) - \psi\left(\frac{x}{6^2} \right) & \le B \frac{x}{6} + 4 \log \frac{x}{6} - 5\\ \psi\left(\frac{x}{6^2} \right) - \psi\left(\frac{x}{6^3} \right) & \le B \frac{x}{6^2} + 4 \log \frac{x}{6^2} - 5\\ \cdots\\ \psi\left(\frac{x}{6^m} \right) - \psi\left(\frac{x}{6^{m+1}} \right) & \le B \frac{x}{6^m} + 4 \log \frac{x}{6^m} - 5 \end{align*} Adding the inequalities gives \begin{equation*} \psi(x) \le \psi\left(\frac{x}{6^{m+1}} \right) + \frac{6}{5} B + 4 (m+1) \log x - (m+1) 5 \end{equation*} The inequality \( 6^{m+1} \le x \) gives \( m + 1 \le \frac{\log x }{\log 6 } \). Also, \( 1 \le \frac{x}{6^{m+1}} < 6 \), thus \( \psi\left(\frac{x}{6^{m+1}} \right) \le 3 \). Therefore, for all \( x \ge 6 \) one has \begin{equation}\label{20150923.8} \psi(x)\le \frac{6}{5} B x + 4 \frac{\log^2 x}{\log 6} \end{equation} One can verify in Mathematica that this inequality is also valid for \( x \ge 1 \)

__Calculation of lower bound on \( \psi(x) \)__

One has \begin{equation*} -2 \psi\left( \frac{x}{6} \right)+\psi\left( \frac{x}{7} \right)+\psi\left( \frac{x}{11} \right) \le 0 \end{equation*} and \begin{equation*} -2 \psi\left( \frac{x}{12} \right)+\psi\left( \frac{x}{13} \right)+\psi\left( \frac{x}{17} \right) \le 0 \end{equation*} and so on. Therefore from \eqref{20150923.2} one obtains \begin{equation*} V(x) \le \psi(x) + \psi\left( \frac{x}{5} \right)\end{equation*} Combining this inequality with \eqref{20150923.5} and \eqref{20150923.8} then gives for \( x \ge 5 \) \begin{equation*} B x - 4 \log x -2 + 2 \log 6 - \frac{6}{5} B \frac{x}{5} - 4 \frac{\log^2 x/5}{\log 6} \le \psi(x) \end{equation*} I simplify this expression slightly by using \( -2 + 2 \log 6 > 0 \), then \begin{equation}\label{20150923.9} \frac{19}{25} B x - 4 \log x - 4 \frac{\log^2 x/5}{\log 6} \le \psi(x) \end{equation}

__Proof of Bertrand's postulate__

Because \( \displaystyle\frac{6/5 B}{19/25 B} < 2 \), the upper and lower bound are sufficiently close and one can now deduce Bertrand's postulate in a standard way. I do not do so here, because the post is already quite long. Details can for example be found in Murty's book.

__Other posts about Bertrand's postulate__

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