I've been working on a problem for a couple of days now and I wanted to see if anyone here had an idea whether this was already proven or where I could find some guidance. I feel this problem is connected to the multinomial theorem but the multinomial theorem is not really what I need . Perhaps someone on this forum knows how this problem would be named and even how to solve it. I feel like this should be easy to prove and I've proven single specific cases but I can't wrap my head around the more general case and in fact have no idea whether it's provable, proven, or even true. So we have a set ω containing n numbers referred to as P1, P2, P3... Each of these numbers is a distinct prime number greater than 4. It may or may not be relevant to the proof, but ω always contains all qualifying prime numbers up to a certain number, so it would be impossible to have missing prime numbers, and the following set would be impossible (ω = P1:3 = [5, 7, 13]). You can only have ω that include all prime numbers up to a point, such as (ω = P1:3 = [5, 7, 11]), or (ω = P1:2 = [5, 7]). Note that because all elements of ω are ordered and distinct, and that 5 is always present in the set, we can write them as prime gaps being added to 5 in the following way: P1 = P1 = 5 P2 = P1 + PG2 = 5 + 2 = 7 P3 = P1 + PG2 + PG3 = 5 + 2 + 4 = 11 ... where PGi reads "the prime gap associated with Pi" and where the indexing of the gaps starts at P1 = 5, and PG1 = 0. So to recapitulate the conditions in a cleaner list: Conditions ω contains all distinct prime numbers between 4 and a given integer (condition 1) ω has a certain size n > 1 (condition 2) Due to Bertrand's postulate, note that any PGi < Pi. Specific formulation of the problem (this is an equivalent of the general case with m = n) Prove that for any given ω with number of elements n > 1, the product of all P in ω is greater than the sum of the products of all possible combinations of subsets of ω with size n - 1. So, for instance, for n = 2, prove that: P1P2 > P1 + P2 or, for another instance with n = 4, prove that: P1P2P3P4 > P1P2P3 + P1P2P4 + P1P3P4 + P2P3P4 General formulation of the problem Prove that for any given ω with number of elements n > 1, given any chosen integer 1 > m ≤ n, the sum of the products of all possible combinations of subsets of ω with size m is greater than the sum of the products of all possible combinations of subsets of ω with size m - 1. So, for instance: for n = 4 and m = 3, P1P2P3 + P1P2P4 + P1P3P4 + P2P3P4 > P1P2 + P1P3 + P1P4 + P2P3 + P2P4 + P3P4 or for n = 5 and m = 4, P1P2P3P4 + P1P2P3P5 + P1P2P4P5 + P1P3P4P5 + P2P3P4P5 > P1P2P3 + P1P2P4 + P1P2P5 + P1P3P4 + P1P3P5 + P1P4P5 + P2P3P4 + P2P3P5 + P2P4P5 + P3P4P5 Example proof I've had success handling single cases by expressing all primes as being an addition of prime gaps with 5, for instance: Proof for m = 2 and n = 2 Prove that P1P2 > P1 + P2 We know that P2 = P1 + PG2, therefore: P1(P1 + PG2) > P1 + P1 + PG2 P12 + P1PG2 > 2P1 + PG2 We may label these 4 terms as term 1, 2, 3 and 4. It is easy to see that term 1 is greater than term 3, because P1 > 2 (condition 1). It is also easy to see that term 2 is greater than term 4, because P1 > 1 (condition 1). Therefore the left side has to be greater than the right side. Help needed! I'd love to see if anyone has some guidance to provide on proving the general case. It feels like this is strongly related to the Pascal triangle and multinomial theorem but I have difficulties finding where to start.