Can one prove by elementary means (such as Paul Erdös' proof of Bertrand's Postulate) that every prime greater than 5 is less than the sum of the two primes immediately preceding it?



Is every prime greater than 5, less than the sum of the two previous primes?
I am to elaborate a bit on how it is that the answer I left in [this previous](https://mathoverflow.net/q/113840/1593) question implies a positive answer to Mr. Recamán's question.


By elementary means (i.e., without resorting to any complex analysis), in his paper [*Mémoire sur les nombres premiers*](http://sites.mathdoc.fr/JMPA/PDF/JMPA_1852_1_17_A19_0.pdf) (Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.), P. L. Chebyshev established that for any $\varepsilon >\frac{1}{5}$, there exists an $n\_{\varepsilon}\in \mathbb{N}$ such that for all $n\geq n\_{\varepsilon}$,


$$\pi((1+\varepsilon)n)−\pi(n)>0.$$


In particular, taking $\varepsilon = \frac{1}{4}$, this implies that, for every sufficiently large $k \in \mathbb{N}$, we have that


$$p\_{k+2} \leq (1.25)^{2}p\_{k} < 2p\_{k}$$


and whence, for every sufficiently large $k \in \mathbb{N}$, it is certainly the case that


$$p\_{k+2} < 2p\_{k} < p\_{k}+p\_{k+1}.$$


By *retracing* Chebyshev's arguments it is possible to obtain an explicit version of the previous conclusion.


