Showing that there is no better $C$ than the optimal is surely not suitable for juniors (needs a limit argument), but otherwise the problem is decent and out of the ordinary boring inequalities. Observe that $a_{n+1} = (a_n + 1)^3 - 1$. Hence if $a_{2020} = t$, then $a_{2021} = (t+1)^3$, $a_{2022} = (t+1)^9$ and $a_{2023} = (t+1)^{27}$. Set $x = t+1 > 1$, then we study \[ \frac{x^{27} - x}{x^9 - x^3} > C.\] The left hand side is $\frac{x^{26} - 1}{x^2(x^6-1)} = \frac{x^{24} + x^{22} + \cdots + x^2 + 1}{x^2(x^4+x^2+1)}$ and now $x\to 1$ yields $C \leq \frac{13}{3}$. To show that $C = \frac{13}{3}$, one can either devise AM-GM combinations or just expand and factor by Horner's method, getting the equivalent \[(x - 1) (x + 1) (3 x^{22} + 6 x^{20} + 9 x^{18} + 12 x^{16} + 15 x^{14} + 18 x^{12} + 21 x^{10} + 24 x^8 + 27 x^6 + 17 x^4 + 7 x^2 - 3) > 0\]or just (no need to bother noticing that $(-1)$ is a root) \[(x-1)(3 x^{23} + 3 x^{22} + 6 x^{21} + 6 x^{20} + 9 x^{19} + 9 x^{18} + 12 x^{17} + 12 x^{16} + 15 x^{15} + 15 x^{14} + 18 x^{13} + 18 x^{12} + 21 x^{11} + 21 x^{10} + 24 x^9 + 24 x^8 + 27 x^7 + 27 x^6 + 17 x^5 + 17 x^4 + 7 x^3 + 7 x^2 - 3 x - 3) > 0.\]All factors are positive for $x>1$, as $7x^2 - 3x - 3 = x^2 + 3x(x-1) + 3(x-1)(x+1) > x^2 > 0$.