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Superabundant Numbers

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9/9/2010
Let nn be a positive integer. Let σ(n)\sigma(n) be the sum of the natural divisors dd of nn (including 11 and nn). We say that an integer m≥1m \geq 1 is superabundant (P.Erdos, 19441944) if ∀k∈{1,2,…,m−1}\forall k \in \{1, 2, \dots , m - 1 \}, σ(m)m>σ(k)k.\frac{\sigma(m)}{m} >\frac{\sigma(k)}{k}. Prove that there exists an infinity of superabundant numbers.
number theoryDivisorssum of divisorsInequalityIMO Shortlist