Show that the inequality holds for all positive integers
We are not going to prove that $a_k\geq 3a_{k-1}$ because it is not true: $$a_3=2a_2+3a_1=2\cdot 9+3\cdot 2=24&lt;3\cdot 9=3a_2$$

We also don't have $a_k&lt;3^k$ for all $k$, since $a_2=9=3^2$. That's probably a typo.

Let's prove, by induction, that for all natural $k$ we have $a_k\leq 3^k$.
We have that $a_1=2\leq 3=3^1$.
Assume that $a_{k-2}\leq 3^{k-2}$ and $a_{k-1}\leq 3^{k-1}$. Or you can assume that $a_t\leq 3^t$ for all $t&lt;k$.
Then
$$\begin{align}
a_k&amp;=2a_{k-1}+3a_{k-2}\\
&amp;\leq 2\cdot 3^{k-1}+3\cdot 3^{k-2}\\
&amp;=3\cdot 3^{k-1}\\
&amp;=3^{k}
\end{align}$$
Therefore, $a_{k}\leq 3^{k}$.
By induction, for all natural $k$ we have that $a_{k}\leq 3^k$.