Prove that for $n\geq3$, $n^3\geq3n+5$ 
Prove that for $n\geq3$,$$n^3\geq3n+5$$

My try:
Let $f(x)=x^3-3x-5$. Clearly $x=1$ and $x=-1$ are critical points where $x=1$ is the local minima so the function $f(x)$ is an increasing function for $x>1$ and the result follows.
Please check and if possible provide another way to solve like induction. Thank you.

Using induction:
$1)$ For $n=3$ we get $3^3\ge 3\cdot 3+5=14$
$2)$ Hypothesis: $n^3\ge 3n+5$
$3)$ For $n+1$
$$(n+1)^3=n^3+(3n^2+3n+1)\ge 3n+5+(3n^2+3n+1)\\
(n+1)^3\ge [3(n+1)+5]+(3n^2+3n-2)\quad (1)$$
but 
$$3n^2+3n-2=3n(n+1)-2\ge0\text{ for } n\ge 3$$
then from $(1)$ we get
$$(n+1)^3\ge 3(n+1)+5$$
