For consecutive primes $a\lt b\lt c$, prove that $a+b\ge c$. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:33:16Z http://mathoverflow.net/feeds/question/113840 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/113840/for-consecutive-primes-a-lt-b-lt-c-prove-that-ab-ge-c For consecutive primes $a\lt b\lt c$, prove that $a+b\ge c$. Bavid 2012-11-19T16:12:47Z 2013-02-10T00:54:54Z <p>For consecutive primes $a\lt b\lt c$, prove that $a+b\ge c$.</p> <p>I cannot find a counter-example to this. Do we know if this inequality is true? Alternatively, is this some documented problem (solved or unsolved)? </p> http://mathoverflow.net/questions/113840/for-consecutive-primes-a-lt-b-lt-c-prove-that-ab-ge-c/113843#113843 Answer by Tony Huynh for For consecutive primes $a\lt b\lt c$, prove that $a+b\ge c$. Tony Huynh 2012-11-19T16:27:01Z 2012-11-19T16:27:01Z <p>Yes, this is true. In 1952, Nagura proved that for $n \geq 25$, there is always a prime between $n$ and $(6/5)n$. Thus, let $p_k$ be a prime at least 25. Then $p_k+p_{k+1} > 2p_k$. But by Nagura's result we have that $p_{k+2} \leq 36/25 p_k &lt; 2p_k$. It is easy to verify the conjecture for small values of $p$. </p> http://mathoverflow.net/questions/113840/for-consecutive-primes-a-lt-b-lt-c-prove-that-ab-ge-c/113882#113882 Answer by quid for For consecutive primes $a\lt b\lt c$, prove that $a+b\ge c$. quid 2012-11-19T23:52:26Z 2012-11-19T23:52:26Z <p><a href="http://www.imsc.res.in/~rao/ramanujan/CamUnivCpapers/Cpaper24/page2.htm" rel="nofollow">Ramanujan (1919), see Eq. (18):</a> </p> <blockquote> <p>$$\pi(x) - \pi(x/2) \ge 2 \quad \text{ for } x\ge 11$$</p> </blockquote> <p>Whence, with $x= 2p_k$ for $p_k \ge 7$, $$p_{k+2} \le 2 p_k \lt p_k+p_{k+1},$$ and $5\le 2+3$, $7\le 3+ 5$, $11 \le 5+7$. </p> http://mathoverflow.net/questions/113840/for-consecutive-primes-a-lt-b-lt-c-prove-that-ab-ge-c/113885#113885 Answer by J. H. S. for For consecutive primes $a\lt b\lt c$, prove that $a+b\ge c$. J. H. S. 2012-11-20T00:40:51Z 2012-11-20T01:07:24Z <p>As a matter of fact, P. L. Chebyshev knew already that for any $\epsilon > \frac{1}{5}$, there exists an $n(\epsilon) \in \mathbb{N}$ such that for all $n\geq n(\epsilon),$ </p> <p>$\pi((1+\epsilon)n)-\pi(n)>0.$</p> <p>In [2], one can find a short report on the problem of determining the smallest $n(\epsilon)$ explicitly once that $\epsilon$ has been fixed.</p> <p><strong>References</strong></p> <p>[1] P. L. Chebyshev. <em>Mémoire sur les nombres premiers</em>. Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.</p> <p>[2] H. Harborth &amp; A. Kemnitz. <em>Calculations for Bertrand's Postulate</em>. Mathematics Magazine, <strong>54</strong> (1), pp. 33-34.</p>