Hi How can be proved by using chebyshev's theorem(?, which theorem?) that for consecutive primes we have
(1+1/p^2)(1+1/q^2)<(1+1/r)
where p,q,r are consecutive primes and greater than 11
thanks
Hi How can be proved by using chebyshev's theorem(?, which theorem?) that for consecutive primes we have (1+1/p^2)(1+1/q^2)<(1+1/r) where p,q,r are consecutive primes and greater than 11 thanks 

closed as too localized by Felipe Voloch, Pete L. Clark, Andrés E. Caicedo, Yemon Choi, José FigueroaO'Farrill Nov 29 '10 at 21:01This 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. 


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, $8p^2+4p^3<0$ which is true for $p\ge 11$. 

