Hint $\ $ Both $\:3^n$ and $\rm\:m_n$ are solutions of $\rm\:f_{n+2} = 2\, f_{n+1} + 3\, f_n,\:$ therefore, by linearity, their difference $\rm\:f_n = 3^n\!-m_n\:$ is also a solution. But it is straightforward to prove by induction that $\rm\:f_1,f_2\ge 0\:$ $\Rightarrow$ $\rm\:f_n \ge 0,\:$ for all $\rm\:n\ge 1,\:$ because $\rm\:f_n,f_{n+1} \ge 0\:$ $\Rightarrow$ $\rm\:f_{n+2} = 2\,f_{n+1}+3\,f_n\ge 0.\:$ Therefore $\rm\: f_n = 3^n\! - m_n \ge 0,\:$ so $\rm\:3^n \ge m_n\:$
Remark $\ $ Note how reformulating it this way makes the essence of the matter obvious, viz. if the recurrence is an increasing function of the prior values, then if the sequence starts with initial values $\ge 0$ then it must remain $\ge 0$ for all greater values, since each successive step is increasing.