use more inequalities..

1.copying... there are two possible prime factorizations for n: n=p^5 or

n=p*q^2.

2. If n=p^5, 1+p^5=5(p+p^2+p^3+p^4).

   p^5 > 5p^4, so p>5, then (p ≥ 7), but 5(p+p^2+p^3+p^4) ≤ 5p^4((1/7)^3 + （1/7)^2 + 1/7 + 1）≤ 5p^4(9/7) ≤ 7p^4 < 1+ p^5.  

2, if n=pq^2, then 1+pq^2 = 5(p+q+pq + q^2). Note that 5(p+q)q < 5(p+q+pq+q^2) -1 = pq^2, p>5, q>5, so p>=7, q>=7.  pq^2 < 5(p+q+pq+q^2) = 5(p+q)(1+q) < (40/7)(p+q)q, so 5(p+q) < pq < (p+q)40/7, 5(1/p + 1/q) < 1 < (1/p + 1/q)40/7. If min(p, q) > 7, then (1/p + 1/q)40/7 <= (1/11 + 1/13)40/7 < 1. So min(p, q) = 7.