Maths Olympiad Prep

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, 1992

Algebra Difficulty 4.8 AIME Prove it Baltic Way

Problem:

Prove that the product of the 99 numbers of the form k3−1k3+1\frac{k^{3}-1}{k^{3}+1} where k=2,3,…,100k=2,3, \ldots, 100, is greater than 23\frac{2}{3}.

Solution

Solution:

Note that

k3−1k3+1=(k−1)(k2+k+1)(k+1)(k2−k+1)=(k−1)(k2+k+1)(k+1)((k−1)2+(k−1)+1) \frac{k^{3}-1}{k^{3}+1}=\frac{(k-1)\left(k^{2}+k+1\right)}{(k+1)\left(k^{2}-k+1\right)}=\frac{(k-1)\left(k^{2}+k+1\right)}{(k+1)\left((k-1)^{2}+(k-1)+1\right)}

After obvious cancellations we get
∏k=2100k3−1k3+1=1⋅2⋅(1002+100+1)100⋅101⋅(12+1+1)>23 \prod_{k=2}^{100} \frac{k^{3}-1}{k^{3}+1}=\frac{1 \cdot 2 \cdot\left(100^{2}+100+1\right)}{100 \cdot 101 \cdot\left(1^{2}+1+1\right)}>\frac{2}{3}

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