MHB How Do You Prove the Inequality Involving Primes and Harmonic Sum for POTW #337?

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The discussion centers on proving the inequality involving primes and the harmonic sum for the Problem of the Week (POTW) #337. The problem states that for primes \( q_1, \ldots, q_r \) in the interval \([1, n]\), the product of \( \left(1 - \frac{1}{q_j}\right) \) multiplied by the harmonic sum \( \sum_{k = 1}^n \frac{1}{k} \) is less than 1. Despite the challenge, no participants provided a solution, indicating a potential difficulty in tackling the problem. A solution is mentioned to be available below the main discussion. The thread emphasizes the importance of engaging with the problem and encourages readers to explore the provided links for further guidance.
Euge
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Here is this week's POTW:

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Suppose $q_1,\ldots, q_r$ are the primes in the interval $[1, n]$ where $n$ is an integer $> 1$. Prove

$$\prod_{j = 1}^r \left(1 - \frac{1}{q_j}\right)\sum_{k = 1}^n \frac{1}{k} < 1$$-----

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No one answered this week's problem. You can read my solution below.
Note $$\prod_{j = 1}^r \frac{1}{1 - \frac{1}{q_j}} = \sum_{m_1 \ge 0} \sum_{m_2 \ge 0}\cdots \sum_{m_r \ge 0} \frac{1}{q_1^{m_1}q_2^{m_2} \cdots q_r^{m_r}} = \sum_{k\in S} \frac{1}{k}$$ where $S$ is the set of all positive integers $k$ whose prime factors are no greater than $n$. Since $$\sum_{k\in S} \frac{1}{k} > \sum_{k = 1}^n \frac{1}{k}$$it follows that $$\prod_{j = 1}^r \frac{1}{1 - \frac{1}{q_j}} > \sum_{k = 1}^n \frac{1}{k}$$ and therefore $$\prod_{j = 1}^r \left(1 - \frac{1}{q_j}\right) \sum_{k = 1}^r \frac{1}{k} < 1$$ as desired.