# A proof of the inequality $latex \varepsilon_{m}\le 2\varepsilon_{moy}$

Given $n\ge 14$ a positive integer, let’s call the difference between two consecutive potential typical primality radii of $n$ a « Goldbach gap ». There are exactly $N_{1}(n)$ such Goldbach gaps. Let’s denote by $\eta(n)$ the number of distinct Goldbach gaps, and let’s write $\varepsilon_{i}$, $1\le i\le \eta(n)$ the $i$-th such Goldbach gap in the increasing order. Thus $\varepsilon_{1}=\varepsilon_{min}$ and $\varepsilon_{\eta(n)}=\varepsilon_{max}$. Moreover, let’s denote by $w_{i}$ the multiplicity of $\varepsilon_{i}$. One has $\displaystyle{\sum_{i}^{\eta(n)}w_{i}\varepsilon_{i}=P_{ord_{C}(n)}}$ and $\displaystyle{\sum_{i}^{\eta(n)}w_{i}=N_{1}(n)}$. Writing $p_{i}:=\dfrac{w_{i}}{N_{1}(n)}$, one gets $\varepsilon_{moy}=\displaystyle{\sum_{i=1}^{\eta(n)}p_{i}\varepsilon_{i}}$. Let’s define $s_{i}$ for $i$ ranging from $1$ to $\eta(n)$ so that $s_{1}=\dfrac{1}{2}$, $s_{\eta(n)}=\dfrac{1}{2}$ and $s_{i}=0$ if $i\not\in\{1,\eta(n)\}$. One has $\vert\varepsilon_{moy}-\varepsilon_{m}\vert=\vert\displaystyle{\sum_{i}^{\eta(n)}\varepsilon_{i}(p_{i}-s_{i})\vert}=\vert(p_{1}-\dfrac{1}{2})\varepsilon_{1}+(p_{\eta(n)}-\dfrac{1}{2})\varepsilon_{\eta(n)}+\displaystyle{\sum_{i=2}^{\eta(n)-1}p_{i}\varepsilon_{i}}\vert$. Hence $\vert\varepsilon_{moy}-\varepsilon_{m}\vert\le\displaystyle{\sum_{i=2}^{\eta(n)-1}p_{i}\varepsilon_{i}}$, and $\vert\varepsilon_{moy}-\varepsilon_{m}\vert\le \varepsilon_{moy}$. Therefore $\varepsilon_{m}\le 2\varepsilon_{moy}$.

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# 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.