Not sure where to go with this
$2^k > k^3$ for $k > 9$
$2^(k+1) > (k+1)^3$
$2^(k+1) = 2^k \cdot 2$
$2^k \cdot 2 > k^3 \cdot 2$ (by inductive hypothesis)
$2^k \cdot 2 > 2k^3$
$2k^3 > k^3 + 3k^2 + 3k + 1$
$2k^3 > (k+1)^3$
I know the final inequality is true, since I graphed it, but I was wondering if there was a clearer way to show the thought process algebraically.
You have $2^{k+1}=2\cdot2^k>2k^3$, so it suffices to show that $2k^3\ge(k+1)^3$, or, equivalently, that $k^3\ge 3k^2+3k+1$ for $k>9$. And
$$3k^2+3k+1<3\left(k^2+k+1\right)\le3\left(3k^2\right)=9k^2<k^3$$
for $k>9$.
You want to show that $2^k > k^3$ implies $2^{k+1} > (k+1)^3$ for $k > 9$.
Let's try proof by contradiction.
Suppose $2^k > k^3$ and $2^{k+1} \le (k+1)^3$. Then $(k+1)^3 \ge 2^{k+1} = 2\ 2^k > 2\ k^3 $ or $k^3 < 3k^2+3k+1$ which, by a non-coincidence, is the same as Brian M. Scott's inequality, and can be shown false the same way for $k > 9$.
