<s>[INST] Prove $3^n \ge n^3$ by induction Yep, prove $3^n \ge n^3$, $n \in \mathbb{N}$.
I can do this myself, but can't figure out any kind of "beautiful" way to do it. 
The way I do it is:
Assume $3^n \ge n^3$
Now, 
$(n+1)^3 = n^3 + 3n^2 + 3n + 1$,
and $\forall{} n \ge 3$,
$3n^2 \le n^3, \,\, 3n + 1 \le n^3$ 
Which finally gives $(n+1)^3 \le 3n^3 \le 3^{n+1}$ by our assumption.
Now just test by hand for n=1,2,3 and the rest follows by induction.
Anyone got anything simpler?
 [/INST] Basis:
$n = 1$
$$3^1 \ge 1^3 \implies 3 \ge 1\text{, which is true}$$
Inductive hypothesis
Let $n=k$ and also let this inequality hold:
$$3^k \ge k^3$$
Inductive step
We'll prove that it also holds when $n=k+1$
$$3^{k+1} \ge (k+1)^3$$
$$3^k \cdot 3 \ge k^3 + 3k^2 + 3k + 1$$
$$3^k + 3^k + 3^k \ge k^3 + 3k^2 + 3k + 1$$
Now obviously as $3^k \ge k^3$ by the inductive hypothesis it's enough to show that
$$2 \cdot 3^k \ge 3k^2 + 3k + 1$$
Note that for every $k \ge 2$, this inequality holds $3k^2 \ge 3k + 1$
$$2 \cdot 3^k \ge 6k^2$$
$$3^k \ge 3k^2$$
This is true for every $k \ge 3$
Now you can prove the other 2 cases $n = 1,2$ to give a complete proof.
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