Dependency of the properties of numbers' subsets

Question

Let $a$ be the number of different sums of all the subsets of a set $A$ of $n$ real numbers (let's suppose the sum of an empty set is $0$). Let $b$ be the number of ordered pairs of the subsets of $A$ with the same sums (the subsets in the pair may be the same one, but they are in that case only one pair). Prove that $6^n\ge ab$

I don't know if it will have any use but I figured out that the number of subsets is $2^n$, $2^n\ge a\ge 1$, $(2^n\cdot(2^n-1))\ge b\ge2^n$. $b$ for a subset of subsets with the same sum is increasing faster than the number of subsets inside this subset. Experimenting I only managed to receive sums of $2$s with different exponents.

Answer 1

As I read it, it is not true. There are $2^n$ subsets of $A$. We can arrange that all subsets have different sums, for example by taking the elements of $A$ to be distinct powers of $2$. Then $a=2^n=b$ and $ab=2^{2n}=4^n \lt 6^n$. I think you want the inequality in the other direction.