Question

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

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 $p\ge 11$.