number theory - Conditions for which $(sum p_i)(sum frac{1}{p_i})$ over an arbitrary $i$ for a set of primes ${p_i}$ is unique?

I am looking for conditions (if any are needed beyond properties of primes) for which $(\sum p_i)(\sum \frac{1}{p_i})$ over an arbitrary $i$ for a set of primes $\{p_i\}$ is unique in that there is no other set of primes $p_j$ for which

$$(\sum p_i)(\sum \frac{1}{p_i}) = (\sum p_j)(\sum \frac{1}{p_j})$$

It is a bit difficult to describe so I will give an example:

I would like to know if there is a proof (if it is even true) that given any primes $p_1$ and $p_2$ that there are no other set of primes $\{p_3$,$p_4,..,p_k\}$ such that

$$(p_1+p_2)(\frac{1}{p_1}+\frac{1}{p_2}) = (p_3+p_4+..+p_k)(\frac{1}{p_3}+\frac{1}{p_4}+..+\frac{1}{p_k})$$

And in general (again I don't know if this is true), that

$$(p_1+p_2+..+p_n)(\frac{1}{p_1}+\frac{1}{p_2}+..+\frac{1}{p_n}) \neq (p_3+p_4+..+p_s)(\frac{1}{p_3}+\frac{1}{p_4}+..+\frac{1}{p_s})$$

So basically whether or not I can find a set of two sets of primes of any length which satisfy the equality.

I have been trying to figure out a clever way to reduce this and would guess that there is some trick by the uniqueness of primes, but I don't see it.

Thanks,

Brian