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.

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