Mathematial proof of sums

Is there a mathematical proof to show that:

if two sets of 7 prime numbers sum to 100 then, then the set that has the largest product of all prime numbers will be the one with the smallest range (highest score minus lowest score).

Thanks

Try this for two numbers, and see if you can take it from there:

Suppose we have two numbers a and b such that:

a+b = X
ab = Y


Now consider what happens if we adjust these numbers by a small positive amount d, making a slightly larger and b slightly smaller:

(a+d) + (b-d) = X
(a+d)(b-d) = ab - ad + bd -d²
= ab - d(a - b + d)
= Y - d(a - b + d)

Now, considering that we have already defined d as a positive amount, then the new product is smaller than the old one as long as a > b. And if a > b, then adding to a and subtracting from b moves them further apart. The conclusion is that if you move the two numbers further apart, then the new product is smaller.



Also consider what happens if a < b. in this case, the new product is larger than before, but since we are adding to a and subtracting from b we have moved them closer together. Again, by moving a and b closer together we get a larger product.

Incidentally, regarding the last point, this is only strictly true if a < b and the difference is less than d. But if this is not the case, what we have done is taken two numbers close together and adjusted them such that the lower one is now the higher and vice versa. But since addition and multiplication are commutative (i.e. a+b = b+a) we might as well have just adjusted the lower one down and the higher one up, i.e. the first scenario.


Now all you have to do is extend this case for two numbers up to 7...


zaza