Proving $n \leq 3^{n/3}$ for $n \geq 0$ via the Well-Ordering Principle I'm attempting to prove:
$$n \leq 3^{n/3} \quad \text{for }n \geq 0$$
I'm having a little trouble continuing. This is what I have so far:

Suppose for a contradiction there is a subset of nonnegative integers $S$ such that $x > 3^{x/3}$ for $x \in S$. By the Well-Ordering Principle, there is some least element $m \in S$. It also means that for some $n < m$, $n \leq 3^{n/3}$ must apply, and since $n = 0$ holds we can conclude that $m > 0$. If follows that $m - 1 \geq 0$ and so $m - 1 \leq 3^{(m-1)/3}$ applies:
$$
\begin{align}
m - 1 \leq 3^{(m-1)/3} &\equiv (m-1)^3 \leq 3^{m-1} \\
&\equiv 3(m-1)^3 \leq 3^m \\
&\equiv 3(m-1)^3 \leq 3^m < m \\
&\equiv 3(m-1)^3 < m
\end{align}
$$

But I'm not sure how to show now that this is a contradiction. How do I continue?
 A: Using contradiction, if there exist a $n \ge 0 $ such that
$$
\begin{align}
n &> 3^{n/3}
\end{align}
$$
Then, because of the WOP, there will be a $n=m$ which is the least value of the set that satisfies the inequality above.
It is verifiable that $n \le 3^{n/3}$ holds true for $0 \le n \le 4$. Thus, $m \ge 5$
since $m-3 \lt m$
$$
\begin{align}
(m-3) &\le 3^{(m-3)/3}
\end{align}
$$
Thus,
$$
\begin{align}
(m-3) &\le 3^{m/3}\cdot3^{-1}\\
3(m-3) &\le 3^{m/3}
\end{align}
$$
It is known that $m \lt 3(m-3)$ for any $m \ge 5$, thus
$$
\begin{align}
 m \lt 3(m-3) &\le 3^{m/3}\\
 m &\lt 3^{m/3}
\end{align}
$$
Which contradicts the premise $n > 3^{n/3}$.
A: From
$3(m-1)^3 < m$
we get
$3m^3-9m^2+8m-3 < 0$
or
$m(3m^2-9m+8) < 3
$
or
$m(3m(m-3)+8) < 3
$.
This is false for
$m \ge 4$
since
$3m(m-3) \ge 3m \ge 12$.
It is also false for $m=3$
by direct computation.
For a more general
non-inductive proof,
you can use the fact that
$x^{1/x}$ is decreasing for
$x \ge e$
so
$e \le a < b$
implies that
$a^{1/a} > b^{1/b}$
or
$a^b > b^a$
or
$a^{b/a} > b$.
Now put $a = 3$.
A: By some explicit calculation, you can see that the inequality holds for $m \leq 2$ which means that that the minimum value of that set of contradictions, $m$ has to be greater than 2. 
$3(m-1)^{3} < m$, which is implied by your original assumption does not hold for any $m \geq 2$. There is the contradiction
A: Let C be the set C ::= {$ n \in N | n > 3^{n/3} $}
For the purpose of obtaining contradiction we assume that C is not empty
There is the smallest element $n_0 \in C$
$n_0 \geq 4$ since for n = 3 property holds
We have : $n_0 - 1 \leq 3^{{(n_0 - 1)}/3}$
$3(n_0 - 1)^3 \leq 3^{n_0} < n_0^3$
Draw the graph of the equation : $3(n_0 - 1)^3 - n_0^3= 0$. We have that for all $n \geq 4$ $3(n_0 - 1)^3 - n_0^3 > 0 $
$->$ contradicting the fact so C is empty.
