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Let $n\geq a_1 > a_2 > \cdots a_k$ be positive integers such that $\text{lcm}(a_i,a_j)\leq n$. Prove that $ia_i\leq n$ for $i=1,2,\cdots, k$.

How do I begin this? (Please provide hints instead of a complete solution. I don't wish to see the solution right now. It would be best if you provide hints and keep the solution inside a spoiler). Maybe we can consider intervals like $[1,\frac{n}{i}]$ and do something.

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closed as off-topic by Saad, Xander Henderson, user99914, Isaac Browne, Taroccoesbrocco Jul 13 '18 at 5:52

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One easy way of seeing this is, $\displaystyle a_ia_j = \text{lcm}(a_i,a_j) \times \text{gcd}(a_i,a_j)$ then, $$\text{lcm}(a_i,a_j) = \frac{a_ia_j}{\text{gcd}(a_i,a_j)} \ge \frac{a_ia_j}{a_i - a_j},\, \quad \text{for } 1 \le i < j \le k \text{ (why?)}$$

Then, we may try to estimate $\displaystyle \frac{1}{a_{i+1}} - \frac{1}{a_i}$ ...

Spoiler:

Spoiler: (since, $\text{gcd}(a_i,a_j) \vert (a_i - a_j) \implies \text{gcd}(a_i,a_j) \le (a_i - a_j)$). Then, by given condition, $$\frac{1}{a_j} - \frac{1}{a_i} \ge \frac{1}{\text{lcm}(a_i,a_j)} \ge \frac{1}{n}, \forall 1 \le i < j \le k \implies \sum_{i=1}^{j-1} \left(\frac{1}{a_{i+1}} - \frac{1}{a_i}\right) \ge \frac{j-1}{n}$$ Hence, $$\frac{1}{a_j} \ge \frac{j-1}{n} + \frac{1}{a_1} \ge \frac{j}{n}$$ as desired.

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    $\begingroup$ It's better that you hide some parts of your answer to make your answer a hint. The OP wants hints. $\endgroup$ – Batominovski Jul 12 '18 at 18:38
  • $\begingroup$ @Batominovski Thanks for the suggestion! $\endgroup$ – r9m Jul 12 '18 at 18:43
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    $\begingroup$ Ahah, the hint was more than enough to get to the conclusion. $\endgroup$ – Mathejunior Jul 12 '18 at 20:01

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