Let’s first assume that all a,b,c,d are at least 2.

If they were all exactly 2, the triple sum would be 4x2x2x2=32, and the pair sum would be 6x2x2=24, so the triple sum is greater. If we increase any of the numbers, say we change a to 2+e, the triple sum would be increase to 32 + 12e , but the pair sum would only increase to 24 + 6e. In short, the pair sum can never catch up with the triple sum. So our assumption that all numbers were at least 2 is false. Without loss of generality we can let a=1. The equation then becomes:

bc + bd + cd + bcd = b + c + d + bc + bd + cd

And so bcd = b + c + d

Looking for a trio of positive integers whose sum is the same as their product, the only candidate is 1,2,3.

So if our a,b,c,d are (in some order) 1,1,2,3, the triple sum and the pair sum are both equal to 17.