Not a solution, but a possible line of approach:
[sp]Start by noticing that $\dfrac{1989}2 = 995 - \dfrac12$, and $\dfrac1{8961} = \dfrac1{9\cdot995+6}.$ Next, notice that $\dfrac12 = \dfrac13+\dfrac16$. So we can subtract $\dfrac12$ from the left side, and $\dfrac13+\dfrac16$ from the right side, so that the inequality becomes $$\sqrt[3]{\dfrac{2}{1}} +\sqrt[3]{\dfrac{3}{2}} +\ldots +\sqrt[3]{\dfrac{996}{995}} - 995 < \frac19 + \frac1{12} + \frac1{15} + \ldots + \frac1{9\cdot995} + \frac1{9\cdot995+3} + \frac1{9\cdot995+6}.$$ If we write this as $$\sum_{n=1}^{995}\Bigl(\sqrt[3]{\dfrac{n+1}n} - 1\Bigr) < \sum_{n=1}^{995}\Bigl(\frac1{9n} + \frac1{9n+3} + \frac1{9n+6}\Bigr),$$ then it is tempting to think that each term on the left might be less than the corresponding term on the right, in other words $$\sqrt[3]{\dfrac{n+1}n} - 1 < \frac1{9n} + \frac1{9n+3} + \frac1{9n+6}\quad (\text{for all }n\geqslant 1).$$ I believe that is true. It certainly holds for $n=1$, because $\sqrt[3]{2} -1 \approx 0.2599$ and $\frac19 + \frac1{12} + \frac1{15} \approx 0.2611$. It also appears to hold for other values of $n$. But it is a delicate inequality – the difference between the two sides is extremely small – and I have not found any convenient way to prove it.[/sp]
