Show by induction that for all $n \in \mathbb{N}$, $3^n \geq 1+2^n$. I started with the base case of P(1), which does work.
$3^1 = 3 \geq 1+2^1 = 3$
I then assumed P(k), which is
$3^k \geq 1+2^k$.
To prove P(k+1), which is $3^{k+1} \geq 1+2^{k+1}$, I have algebraically manipulated the following:
$3^{k+1} = 3 \cdot 3^k > 3^k \geq 1+2^k$
$2 \cdot 3^k \geq 2 + 2^{k+1}$
$2 \cdot 3^k - 1 \geq 1 + 2^{k+1}$.
At this juncture, I would obviously want to prove that $3^{k+1} > 2 \cdot 3^k - 1$. This is where I have run into trouble. If anyone could either help prove that, or show an alternate way of going about the problem, that would be fantastic.

If $3^k\geqslant1+2^k$, then\begin{align}3^{k+1}&=3\times3^k\\&\geqslant3\times(1+2^k)\\&=3+3\times2^k\\&\geqslant1+2\times2^k\\&=1+2^{k+1}.\end{align}
