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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)?

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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 $c$. – GH from MO Nov 20 '12 at 0:10
(Of course 2.001 can be replaced by 2 unconditionally.) – Charles Feb 10 '13 at 4:13
up vote 37 down vote accepted

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 $p_{k+2} \leq 36/25 p_k < 2p_k$. It is easy to verify the conjecture for small values of $p$.

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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. – Ben Crowell Nov 19 '12 at 21:58
@Ben: I think the statement quoted from 1952 is elementary and can be proved in much the same way as Bertrand-Chebyshev. – GH from MO Nov 20 '12 at 0:04
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. – Steven Sivek Nov 20 '12 at 0:42
@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. – 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 $5\le 2+3$, $7\le 3+ 5$, $11 \le 5+7$.

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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.

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