Induction inequality/ n in exponential I need to find out through induction for which $n \geq 0$ the following inequality holds
$$3n+2^{n} \leq 3^n$$
clearly it does not work for n=1 and n=2, so my induction hypothesis is that it works for $n \geq 3$,
however when performing the induction step, I am kind of stuck
because
$ n \rightarrow n+1$
$3 (n+1) + 2^{n+1} \leq 3^{n+1}$
$3(n+1) + 2^{n} \cdot 2 \leq 3^n \cdot 3$
I  recreated the starting inequality within (overline)
$3n + 3 + 2^n + 2^n \leq (2+1) \cdot 3^n$
$2^n + 3 + \overline{3n + 2^n \leq 3^n} + 2\cdot 3^n$
I tried to show that what was added on the LHS is always smaller than on the RHS, therefor
$2n+3 \leq 2\cdot 3^n$
$2n + 3 \leq 3^n + 3^n$
so it is clear that this holds,but would the proof work that way? if yes, are there other, more elegant approaches?

Another way.
For $n\geq3$ by AM-GM we obtain:
$$3^n-2^n=(3-2)\left(3^{n-1}+3^{n-2}\cdot2+...+3\cdot2^{n-2}+2^{n-1}\right)\geq$$
$$\geq n\sqrt[n]{3^{n-1+n-2+...+1}2^{1+...+n-1+n-2}}=n\cdot6^{\frac{n-1}{2}}\geq n\cdot6^{\frac{3-1}{2}}=6n>3n.$$
