Proof by Induction for Exponential Term Prove by mathematical induction that $3^n\geq n2^n \text{ } \forall n\in \mathbb{N}$ the set of natural numbers.  So after showing that $S(1)$ is true, we assume $S(k)$ is true and try to prove that $S(k+1)$ is true.  I make it this far, and get stuck with further justification:
$$\begin{align}
3^k &\geq k2^k \\
3\times 3^k &\geq 2\times k2^k \text{ since } 3>2 \\
3^{k+1} & \geq k2^{k+1}
\end{align}$$
But now I have to somehow get that $k+1$ term in front of $2^{k+1}$, but I'm not sure how to do it in a way that's justified.  Perhaps I've taken a wrong starting approach.  How would you do this?
 A: At your second line, you are throwing away information. Instead
how about $3^{k+1}\ge 3k2^k=k2^{k+1}+k2^k$? If you can show that $k2^k\ge 2^{k+1}$ then you deduce $3^{k+1}\ge k2^{k+1}+2^{k+1}=(k+1)2^{k+1}$.
A: Assume that $3^k \ge k\cdot 2^k$ for some $k$. So
$3^{k+1}=3\cdot 3^k \ge 3\cdot k\cdot 2^k$.
for $k\ge 2$, you have $3\cdot k = 2\cdot k+k \ge 2\cdot k+2 = 2(k+1)$, so
$3^{k+1} \ge 2\cdot (k+1)\cdot 2^k = (k+1)\cdot2^{k+1}$
but, initially, you must show that S(1) and S(2) is true, because the induction step holds only for $k\ge2$.
A: The inequality is true for $n=2,3 $
We have that:
$$3^n \ge n2^n$$
$$3^{n+1}\ge 3n2^n$$
$$3^{n+1}\ge n2^{n+1}+n2^n$$
It's easy to show that for $n\ge 2$
$$n2^n \ge 2^{n+1}$$
$$n \ge 2$$
Therefore
$$3^{n+1}\ge (n+1)2^{n+1}$$

Another way ...
$$3^n \ge n2^n \implies \left (\frac 32 \right )^n \ge n \implies \left (\frac 32 \right )^{n+1} \ge \frac {3n} 2$$
$$\left (\frac 32 \right )^{n+1} \ge  n+ \frac {n} 2$$
it's easy to show that 
$$n+ \frac {n} 2 \ge n+1$$
$$ \frac {n} 2 \ge 1$$
$$  {n}  \ge 2$$
So it's true for $n \ge 2$ and:
$$\boxed {\left (\frac 32 \right )^{n+1} \ge n+1}$$
