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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

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

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1 Answer

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 $p\ge 11$.

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