# A proof of the upper bound r0(n)=O(log^4 n)

Given a positive integer $n\ge 14$, let $\varepsilon_{min}$ (respectively $\varepsilon_{max}$) be the minimal (respectively maximal) distance between two consecutive potential typical primality radii of $n$. Moreover, let $\varepsilon_{m}:=\dfrac{\varepsilon_{min}+\varepsilon_{max}}{2}$ and $\varepsilon_{moy}:=\dfrac{P_{ord_{C}}(n)}{N_{1}(n)}$. One has $\varepsilon_{m}\le 2\varepsilon_{moy}$ (this inequality will be proved in the next article).

Using the inequality of arithmetic and geometric means, one can write the following inequality: $\varepsilon_{min}.\varepsilon_{max}\le \varepsilon_{m}^{2}$. As  $\varepsilon_{m}\le 2\varepsilon_{moy}$, one gets $\varepsilon_{min}.\varepsilon_{max}\le 4\varepsilon_{moy}^{2}$. So that $\varepsilon_{max}\le\dfrac{4}{\varepsilon_{min}}\left(\dfrac{P_{ord_{C}}(n)}{N_{1}(n)}\right)^{2}$. Obviously $2r_{0}(n)\le \varepsilon_{max}$, hence $r_{0}(n)\le\dfrac{2}{\varepsilon_{min}}\left(\dfrac{P_{ord_{C}}(n)}{N_{1}(n)}\right)^{2}$. As $\varepsilon_{min}\ge 2$, one finally gets $r_{0}(n)\le\left(\dfrac{P_{ord_{C}}(n)}{N_{1}(n)}\right)^{2}\lesssim\dfrac{\log^{4} n}{4.C_{J}^{2}}$, where by $x\lesssim y$ we mean $x\le (1+o(1)) y$ and where $C_{J}=0.66016...$ is the twin prime constant. Hence there exists $n_{0}$ such that $n\ge n_{0}\Rightarrow r_{0}(n)\le n-1$. Therefore every large enough even integer is the sum of two primes.