I am stumped trying to prove that there exists a real number $c$, such that $f(n)\leq cn^4$ for most natural numbers $n$.
$$f(n) = \left\{ \begin{array}{ll} 10, &n=10\\ 3f\left(\left\lfloor \frac{2n}{5} \right\rfloor \right) + 6n^4,&n\geq 1 \end{array} \right.$$
I have computed some values of $f(n)$ from $0-9$ and have chose my constants that seem to make $f(n)\leq cn^4$ true: $c=7$ and $n>4$. However, I don't know how to prove that $c$ is valid. Thanks for your time.
Use (strong) induction:
for $n$ big enough, such as $k< n\implies f(k) \le Ck^4$, $$f(n) = 3f\left(\left\lfloor \frac{2n}{5} \right\rfloor \right) + 6n^4 \le \left[ 3C \frac {2^4}{5^4} + 6\right]n^4 \le \left( 0.08 C + 6\right)n^4 \le Cn^4 $$ as soon as $$ 0.08 C + 6 \le C \Leftarrow C \ge 7 $$ Then check that $7$ is convenient for the first values (until $n > \lfloor \frac{2n}{5} \rfloor $), and you are done.