Showing $4^m>2m^2 + 5m$ for $m\geq 3$ by induction Show that $4^m>2m^2 + 5m$. 
I want to practice proving by induction. I know that this is not true for small values of $m$. So, the base case can be $m=3$. Then we have $$4^3 > 2(3)^2+5(3)$$ = $$64>32$$
Next, we say if $$4^m>2m^2 + 5m$$ then $$4^{m+1}>2(m+1)^2 + 5(m+1)$$
= $$4^m(4) >(m+1)(2(m+1)+1)$$
I am not quite certain what to do with this from here. 
 A: Here is how you can reach case $m+1$ by assuming case $m$:
$$4^m>2m^2 + 5m\\
\Rightarrow4(4^m)>4(2m^2 + 5m)\\
\begin{align}\Rightarrow4^{m+1}&>8m^2+20m\\
&=2m^2+9m+6m^2+11m\\
&>2m^2+9m+7\\
&=2m^2+4m+2+5m+5\\
&=2(m^2+2m+1)+5m+5\\
&=2(m+1)^2+5(m+1)
\end{align}$$
A: You have already established the base case. Now, let $S(m)$ denote the following proposition for all $m\geq 3$:
$$
S(m) : 4^m>2m^2+5m.
$$
Fix some $k\geq 3$ and assume that
$$
S(k) : 4^k>2k^2+5k
$$
holds. To be shown is that $S(k+1)$ follows:
$$
S(k+1) : 4^{k+1} > 2(k+1)^2+5(k+1).
$$
Starting with the left-hand side of $S(k+1)$, 
\begin{align}
4^{k+1} &= 4\cdot 4^k\tag{by definition}\\[0.5em]
  &> 4\cdot(2k^2+5k)\tag{by $S(k)$}\\[0.5em]
  &= 8k^2+20k\tag{expand}\\[0.5em]
  &> 2k^2+9k+7\tag{"strategic observation"}\\[0.5em]
  &= 2(k^2+2k+1)+5k+5\tag{strategically rearrange}\\[0.5em]
  &= 2(k+1)^2+5(k+1),
\end{align}
we see that the right-hand side of $S(k+1)$ follows.
Thus, by mathematical induction, $S(m)$ holds for all $m\geq 3$. $\blacksquare$

I should probably note how I came to my "strategic observation." Essentially, I myself got to the "expand" step and wondered what I might need next. Knowing that I ultimately needed to end up at $2(k+1)^2+5(k+1)$, I expanded everything for this sum and essentially worked backwards, showing what I needed and where. Does it make sense now? 
