Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi How can be proved by using chebyshev's theorem(?, which theorem?) that for consecutive primes we have


where p,q,r are consecutive primes and greater than 11


share|improve this question

closed as too localized by Felipe Voloch, Pete L. Clark, Andres Caicedo, Yemon Choi, José Figueroa-O'Farrill Nov 29 '10 at 21:01

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

add comment

1 Answer 1

Replace $q=p$, the LHS gets bigger. Replace $r=4p$, the RHS gets smaller by the Chebyshev theorem (see Wiki). Now the inequality becomes, after simplification, $8p^2+4-p^3<0$ which is true for $p\ge 11$.

share|improve this answer
add comment

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