Baltic Way competition problems, 1992

Solution:

Note that

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

After obvious cancellations we get
$$\prod_{k=2}^{100} \frac{k^{3}-1}{k^{3}+1}=\frac{1 \cdot 2 \cdot\left(100^{2}+100+1\right)}{100 \cdot 101 \cdot\left(1^{2}+1+1\right)}>\frac{2}{3}$$