Prove that $\prod_{k=2}^{99}\frac{k^{3}-1}{k^{3}+1}$ is greater than $\frac{2}{3}$. I have to prove that the product
$$\prod_{k=2}^{99}\frac{k^{3}-1}{k^{3}+1}$$
is greater than $\displaystyle\frac{2}{3}$.
I've tried to write $k^{3}-1$ as $(k-1)(k^{2}+k+1)$ or another ways but I couldn't finish it.

$\begin{array}\\
f(n)
&=\prod_{k=2}^{n}\dfrac{k^{3}-1}{k^{3}+1}\\
&=\dfrac{\prod_{k=2}^{n}(k^{3}-1)}{\prod_{k=2}^{n}(k^{3}+1)}\\
&=\dfrac{\prod_{k=2}^{n}(k-1)(k^2+k+1)}{\prod_{k=2}^{n}(k+1)(k^2-k+1)}\\
&=\dfrac{\prod_{k=1}^{n-1}k}{\prod_{k=3}^{n+1}k}\dfrac{\prod_{k=3}^{n+1}((k-1)^2+(k-1)+1)}{\prod_{k=2}^{n}(k^2-k+1)}
\qquad\text{(this is the only clever step)}\\
&=\dfrac{2}{n(n+1)}\dfrac{\prod_{k=3}^{n+1}(k^2-2k+1+k-1+1)}{\prod_{k=2}^{n}(k^2-k+1)}\\
&=\dfrac{2}{n(n+1)}\dfrac{\prod_{k=3}^{n+1}(k^2-k+1)}{\prod_{k=2}^{n}(k^2-k+1)}\\
&=\dfrac{2}{n(n+1)}\dfrac{(n+1)^2-(n+1)+1)}{2^2-2+1}\\
&=\dfrac{2}{n(n+1)}\dfrac{n^2+2n+1-n-1+1}{3}\\
&=\dfrac23\dfrac{n^2+n+1}{n^2+n}\\
&=\dfrac23(1+\dfrac1{n^2+n})\\
&\gt \dfrac23
\qquad\text{for all } n\\
\end{array}
$
