For consecutive primes $a\lt b\lt c$, prove that $a+b\ge c$.

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)?

  • 8
    $\begingroup$ Just to complement the responses below: The prime number theorem says that the $n$-th prime is asymptotically $n\log n$, whence your sum $a+b$ is asymptotically $2c$. So your inequality holds for large $c$ without any calculation, in fact $2.001 c>a+b>1.999 c$ for large $c$. $\endgroup$ – GH from MO Nov 20 '12 at 0:10
  • 1
    $\begingroup$ (Of course 2.001 can be replaced by 2 unconditionally.) $\endgroup$ – Charles Feb 10 '13 at 4:13

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 $p_{k+2} \leq 36/25 p_k < 2p_k$. Finally, one can easily check by hand that the result holds for small primes.

  • 1
    $\begingroup$ Hard to believe that math as modern as 1952 is needed in order to prove such an elementary-sounding statement. The 1850 Bertrand–Chebyshev theorem almost, but not quite, does the job. $\endgroup$ – Ben Crowell Nov 19 '12 at 21:58
  • 2
    $\begingroup$ @Ben: I think the statement quoted from 1952 is elementary and can be proved in much the same way as Bertrand-Chebyshev. $\endgroup$ – GH from MO Nov 20 '12 at 0:04
  • 1
    $\begingroup$ You can get away with only using work of Chebyshev for large enough a: Let f(n) = \sum log(p) over all primes p up to n (usually denoted theta(n)). If a+b<c then c>2a and so there's at most one prime between a+1 and 2a, hence f(2a)-f(a) < log(2a). He showed that f(a) < a*log(4), and he proved a bound pi(N) > 0.9N/log(N) for N large, so we should have f(a) >= 0.7a for a large. Then for such a we have log(2a) > f(2a)-f(a) >= (1.4-log(4))a > 0.0137a, which is impossible if a is large in the above sense and at least 505. $\endgroup$ – Steven Sivek Nov 20 '12 at 0:42
  • $\begingroup$ @Steven: Thanks for this argument. I believe Chebyshev proved $f(a)<a *\log 4$ with a better constant than $\log 4$. The factor $\log 4$ comes from Erdős's elegant proof based on $\prod_{n<p<2n}p\leq\binom{2n}{n}$. Apologies in advance if I am wrong here. $\endgroup$ – GH from MO Nov 20 '12 at 1:52

Ramanujan (1919), see Eq. (18):

$$\pi(x) - \pi(x/2) \ge 2 \quad \text{ for } x\ge 11 $$

Whence, with $x= 2p_k$ for $p_k \ge 7$, $$p_{k+2} \le 2 p_k \lt p_k+p_{k+1}, $$ and $5\le 2+3$, $7\le 3+ 5$, $11 \le 5+7$.


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 $n\geq n(\epsilon),$


In [2], one can find a short report on the problem of determining the smallest $n(\epsilon)$ explicitly once that $\epsilon$ has been fixed.


[1] P. L. Chebyshev. Mémoire sur les nombres premiers. Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.

[2] H. Harborth & A. Kemnitz. Calculations for Bertrand's Postulate. Mathematics Magazine, 54 (1), pp. 33-34.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.