start by "crossing numbers off the list". for example:

σ(p) = p+1 < 2p (if p is an odd prime).

σ(pq) = pq + p + q + 1 < pq + p + p + 1 = pq + 2p + 1 ≤ pq + pq = 2pq (when p,q are distinct odd primes with p > q)

(to see the last inequality, note that if q is an odd prime: q - 2 ≥ 1, so p(q - 2) ≥ 1, so pq ≥ 2p + 1).

σ(p^{2}) = p^{2}+ p + 1 < p^{2}+ 2p < p^{2}+ p^{2}= 2p^{2}(for any odd prime p).

can you think of some more ways to eliminate numbers without actually computing σ(n) for every single number?