Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Prove it New Zealand

Problem:

Prove the following inequality
$\frac{6}{2024^{3}} < \left(1 - \frac{3}{4}\right)\left(1 - \frac{3}{5}\right)\left(1 - \frac{3}{6}\right)\left(1 - \frac{3}{7}\right)\dots \left(1 - \frac{3}{2025}\right).$

Solution

Solution:

$\frac{1}{4} \times \frac{2}{5} \times \frac{3}{6} \times \frac{4}{7} \times \frac{5}{8} \times \frac{6}{9} \times \frac{7}{10} \times \dots \times \frac{2022}{2025}$

$= \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times \dots \times 2022}{4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \times \dots \times 2025}$

$= \frac{1 \times 2 \times 3}{2023 \times 2024 \times 2025}$

$= \frac{6}{2024(2024 - 1)(2024 + 1)}$

$= \frac{6}{2024(2024^{2} - 1)}$

$> \frac{6}{2024(2024^{2})}$

$= \frac{6}{2024^{3}}.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.