I want to prove that for all $p_1,..., p_k$, $q_1,...q_l$ distincts prime numbers, this equality is false : $$\left( \sum_{i = 1 }^k \frac{1}{p_i} \right) \left( \sum_{j=1 }^l \frac{1}{q_j} \right) =1$$
I can prove that $k$ and $l$ must be strictly greater than 1, but nothing else. I tried elementary manipulations and the case k=2 but was not able to make any progress.
Edit : For the case k = 2 (and the general case), I tried to write the RHS of $$\frac{1}{p_1} = \frac{1}{\sum_{j} q_j } -\frac{1}{p_2}$$ as an irreducible fraction but was not able to get anything from there. I also tried to find bounds on the RHS. In the case $k=2$, I can prove that $l\geq 3$ (all primes are $\geq 2$ and if $l=2$ we have a sum of 4 terms of the form $\frac{1}{pq}$ on the LHS. Reasoning this way, we can also get $l\geq 4$ and maybe $5$, but it's not really helping) .