Proving $a_{9000}>30$ when $a_1=1$, $a_{n+1}=a_n+ \frac{1} {a_n^2}$ 
A sequence is defined by $a_1=1$, $$a_{n+1}=a_n+\frac1{a_n^2}\;.$$ Show that $a_{9000}>30$.

Source:Problem Solving Strategies by Arthur Engel : Pg 226 , Q23.
 A: Seeing that $a_{n+1} - a_n = a_n^{-2}$, we should try to compare this to the differential equation $y' = y^{-2}$.
This has solution $y = ax^{1/3}$ for some $a$, so this suggests that we should look at $a_n^3$ and see if we can show that it has to grow linearly.
$a_n$ is increasing so it is positive, and then
$a_{n+1}^3 = a_n^3 + 3 + 3a_n^{-3} + a_n^{-6} \ge a_n^3 + 3$. Hence $a_n^3 \ge a_1^3 + 3(n-1)$.
Finally, $a_{9000} \ge (1 + 8999.3)^{1/3}$ which is barely smaller than $30$.
Some more precise calculations should be necessary. Starting from $a_2$ instead of $a_1$ should be enough.
A: Hint: Show that $a_{n+1}^3\gt a_n^3+3$ for every $n$.
Edit: (Upon request by the OP) Thus $a_n^3\geqslant a_2^3+3(n-2)$ for every $n\geqslant2$ and in particular $a_{9000}^3\gt2^3+3\cdot8998=27002\gt30^3$ hence $a_{9000}\gt30$. (Note that the more direct approach starting from $a_n^3\gt a_1^3+3(n-1)$ for every $n\geqslant1$ yields $a_{9000}^3\gt1^3+3\cdot8999=26998$, which just misses the target.) More generally, $a_n^3\geqslant3n+2\gt3n$ hence $a_n\gt\sqrt[3]{3n}$ for every $n\geqslant2$.
