Induction proof with Fibonacci numbers

Prove by induction that for Fibonacci numbers from some index $i > 10$

$1.5^i ≤ f_i ≤ 2^i$

Notice! Because Fibonacci number is a sum of 2 previous Fibonacci numbers, in the induction hypothesis we must assume that the expression holds for k+1 (and in that case also for k) and on the basis of this prove that it also holds for k+2.

This where I've got so far:

Base case: $i = 11$
$f_{11} = 89$
$1.5^{11} ≤ 89 ≤ 2^{11}$ OK!

Induction hypothesis:
$1.5^{k+1} ≤ f_{k+1} ≤ 2^{k+1}$

Induction step:
$1.5^{k+2} ≤ f_{k+2} ≤ 2^{k+2}$

Now I have no idea how to continue from here. Could someone help?

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You forgot to check your second base case: $1.5^{12}\le 144\le 2^{12}$
Now, for your induction step, you must assume that $1.5^k\le f_k\le 2^k$ and that $1.5^{k+1}\le f_{k+1}\le 2^{k+1}.$ We can immediately see, then, that $$f_{k+2}=f_k+f_{k+1}\le 2^k+f_{k+1}\le 2^k+2^{k+1}= 2^k(1+2)\le 2^k\cdot 4=2^{k+2}$$ As for the other inequality, we similarly see that $$f_{k+2}=f_k+f_{k+1}\ge 1.5^k+1.5^{k+1}=1.5^k(1+1.5)=1.5^k\cdot 2.5\ge1.5^k\cdot 2.25=1.5^{k+2}$$
If $\alpha^k\le f_k\le \beta^k$ and $\alpha^{k+1}\le f_{k+1}\le \beta^k$, then $$f_{k+2}=f_k+f_{k+1}\ge \alpha^k+\alpha^{k+1}=\alpha^{k+2}\cdot(\frac1{\alpha^2}+\frac1\alpha)$$ and $$f_{k+2}=f_k+f_{k+1}\le \beta^k+\beta^{k+1}=\beta^{k+2}\cdot(\frac1{\beta^2}+\frac1\beta),$$ so in order to conclude $$\alpha^{k+2}\le f_{k+2}\le \beta^{k+2}$$ is is sufficent to have $\frac1{\alpha^2}+\frac1\alpha\ge 1$ and $\frac1{\beta^2}+\frac1\beta\le 1$. You can verify that this is indeed true for $\alpha=\frac32$ and $\beta=2$.