Let $n$ be a positive integer, and let $k$ be a fixed positive integer. Consider all integers $N_1, N_2, \dots, N_r$ less than or equal to $n$ that are square-free and have exactly $2k$ distinct prime factors. For each $N_j$, choose a factorization $N_j = P_j Q_j$, where $P_j$ and $ Q_j$ are square-free numbers with exactly $k$ distinct prime factors each, satisfying $Q_j > P_j$.
Define the sum:
$$
S = \sum_{j=1}^r N_j.
$$
Conjecture:
It is always possible to choose factorizations of $P_j$ and $Q_j$ such that the following inequality holds: $$ \sqrt{n} \cdot \sum_{j=1}^r P_j \leq S. $$
My Question:
What techniques could be used to rigorously prove the inequality? Specifically:
- Does the strategy of minimizing $P_j$ (to maximize $Q_j$) always ensure that the inequality holds? Note that we have $Q_j > \sqrt{N_j}$ for all $j$, but not necessarily $Q_j > \sqrt{n}$. For instance, consider $n=30$ and $N_j = 2\cdot 3$.
- Are there known probabilistic, combinatorial, or analytic methods that might establish the validity of this conjecture?
Any insights into related problems, similar results, or counterexamples would be greatly appreciated. Thank you!
EDIT: Possible strategy to prove the inequality
Let us fix a choice of factorization for each $N_j = P_j Q_j$, where both $P_j$ and $Q_j$ are square-free and have exactly $k$ distinct prime factors, with $Q_j > P_j$. Suppose we always choose $P_j$ to be the minimum possible among the $k$-factor square-free divisors of $N_j$. Then each such $P_j$ may appear multiple times across different $N_j$, so we group terms by their common $P_j$ values.
Let $P^{(i)}$ be a fixed such value, and let it occur in $s_i$ factorizations. Denote the corresponding $Q_j$'s in that group by $Q_1^{(i)}, \dots, Q_{s_i}^{(i)}$, so that the full expression becomes: $$ S = \sum_{j=1}^r N_j = \sum_i \sum_{j=1}^{s_i} P^{(i)} Q_j^{(i)} = \sum_i P^{(i)} \sum_{j=1}^{s_i} Q_j^{(i)}. $$ Meanwhile, the term we want to compare this to is: $$ \sqrt{n} \cdot \sum_{j=1}^r P_j = \sqrt{n} \cdot \sum_i s_i P^{(i)}. $$ Therefore, the inequality $$ \sqrt{n} \cdot \sum_{j=1}^r P_j \leq \sum_{j=1}^r N_j $$ is equivalent to proving that for each group ( i ): $$ \sum_{j=1}^{s_i} Q_j^{(i)} \geq s_i \sqrt{n}, $$ or equivalently, $$ \frac{1}{s_i} \sum_{j=1}^{s_i} Q_j^{(i)} \geq \sqrt{n}. $$ This reduces the conjecture to the following condition:
For each group of factorizations sharing the same minimal $P^{(i)}$, the average of the corresponding $Q_j = N_j / P^{(i)}$ exceeds $\sqrt{n}$.
This is plausible for the following reasons:
Since $P_j$ is minimized, the corresponding $Q_j = N_j / P_j$ is maximized within each factorization.
Empirically, for large $n$, most $Q_j$ are significantly greater than $\sqrt{n}$, so the average in each group should naturally exceed $\sqrt{n}$.
If this condition on the group averages holds, then the desired inequality follows immediately:
$$ \sqrt{n} \cdot \sum_j P_j \leq \sum_i P^{(i)} \cdot s_i \sqrt{n} \leq \sum_i P^{(i)} \sum_{j=1}^{s_i} Q_j^{(i)} = \sum_j N_j. $$
Thus, the conjecture would be proved.
A natural next step would be to formalize the argument that the average of $Q_j$ over each group exceeds $\sqrt{n}$, possibly using probabilistic or analytic methods over the distribution of square-free integers with a given number of prime divisors. However, achieving that exceeds my current knowledge.
