Strong induction inequality proof using floor

Question

My professor showed this inequality to class today and I've been having a tough time trying to solve it. I know that I have to use strong induction but don't know how to progress farther than the 2 base cases(n=1, n=2).

Recurrence relation: $$F(i) =
\begin{cases}
1, & i \leq 2\\
3 F(\lfloor\frac{i}{3}\rfloor) + 5i, & i > 2
\end{cases}$$

To prove: $ F(n) \leq 4n \cdot log_{2}(n) +1 $

Answer 1

For the induction step assume that F(n) ≤ 4n·log₂(n)+1 for n∈ℕ.
Then, we get for k∈{0,1,2}:
F(3n+k) = 3F(⌊^3n+k/₃⌋)+5(3n+k) = 3F(n)+5(3n+k) ≤ 3(4n·log₂(n)+1)+5(3n+k) = 12n·log₂(n)+3+5(3n+k) ≤ 12n·log₂(n)+6(3n+k) < 4(3n+k)·log₂(n)+4(3n+k)·log₂(3) = 4(3n+k)·log₂(3n) < 4(3n+k)·log₂(3n+k)+1
And this is everything we had to prove.

To the second equal sign:
We have k∈{0,1,2} such that ^k/₃ < 1. So we get
n ≤ n+^k/₃ < n+1
And (because n is a natural number) this inequality leads to
⌊n+^k/₃⌋ = n
And this results in
⌊^3n+k/₃⌋ = ⌊³ⁿ/₃+^k/₃⌋ = ⌊n+^k/₃⌋ = n