Question:
How prove that :$$\prod_{k=2}^n \left({\frac{k^3-1}{k^3+1}} \right) \le\frac{3}{2}$$
My seggestion: we know that : $(k^3+1)=(k+1)(k^2-k+1)$ $(k^3-1)=(k-1)(k^2+k+1)$
This implies that : $$ \prod_{k=2}^n \left({\frac{k^3-1}{k^3+1}} \right) = \prod_{k=2}^n \left({\frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)}} \right) = \prod_{k=2}^n \left({\frac{k-1}{k+1}}\right)\prod_{k=2}^n \left({\frac{k^2+k+1}{k^2-k+1}} \right) $$
and we know that : $$\prod_{k=2}^n \left({\frac{k-1}{k+1}}\right)=\frac {3}{1}\cdot\frac {4}{2}\cdot\frac {5}{3}\cdot\frac {6}{4}\cdot\cdot\cdot\frac {n-2}{n-4}\cdot\frac {n-1}{n-3}\cdot\frac {n}{n-2}\cdot\frac {n＋1}{n-1}＝\frac {n(n＋1)}{2}$$ and therfore : $$\prod_{k=2}^n \left({\frac{k^3-1}{k^3+1}} \right) =\frac{n(n+1)}{2}\prod_{k=2}^n \left({\frac{k^2+k+1}{k^2-k+1}} \right) $$
As Yuumita suggested, $\frac{k^3−1}{k^3+1}<\frac{k^3−1}{k^3+1}+\frac{2}{k^3+1}=1$. Furthermore, this product tends to $\frac{2}{3}$. So if that was what you're actually looking for, there are a few answers that can help you with that.