How do I prove $n! > n^3 , \forall n ≥6, n \in \mathbb{N}$?

Question

$P(n) : n! > n^3 , \forall n ≥6, n \in \mathbb{N}$

I'm not able to get the induction step done. Can anyone help me out with this? I'm stuck on this for the past 30 minutes.

It'd be helpful if you could also let me know your thought process.

Answer 1

Hint: Multiplying $$n!>n^3$$ by $n+1>0$ we get

$$n!(n+1)=(n+1)!>n^3(n+1)$$ it remaines to prove that $$n^3(n+1)>(n+1)^3$$ is hold. The last inequality is equivalent to $$n^3-n^2-2n-1>0$$ Since $$n!=(n-1)!n$$ we can also write $$(n-1)!>n^2$$

Answer 2

The induction step starts with

$$(n+1)!=(n+1)(n!)>(n+1)(n^3)$$

So you have to show that $$(n+1)(n^3)\ge (n+1)^3$$ for $n\ge 6$

That is proving $$(n^3)\ge (n+1)^2$$ for $n\ge 6$

Can you take over from here?

Answer 3

Warning: This is not an inductive argument. However, it is too long to be written as a comment. Other answers have covered how to do the job inductively.

Suppose that $n\geq 6$ is an integer. Let $[n]:=\{1,2,\ldots,n\}$. Define $$S:=\big\{(a,b,c)\big|\,a,b,c\in[n]\big\}=[n]^3$$ and $$T:=\big\{\left(x_1,x_2,\ldots,x_n\right)\,\big|\,\left(x_1,x_2,\ldots,x_n\right)\text{ is a permutation of }(1,2,\ldots,n)\big\}\,.$$ We shall construct an injective function $f:S\to T$ that is not surjective. It follows immediately that $$n^3=|S|=\big|\text{im}(f)\big|<|T|=n!\,.$$

Let $a$, $b$, and $c$ be arbitrary elements of $[n]$. If $a$, $b$, and $c$ are pairwise distinct elements of $[n]$, then let $\left\{x_4,x_5,\ldots,x_n\right\}=[n]\setminus\{a,b,c\}$, where $x_4<x_5<\ldots<x_n$. We define $$f\big((a,b,c)\big):=\left(a,b,c,x_4,x_5,x_6,x_7,x_8,\ldots,x_n\right)\,.$$
If $a=b$ and $b\neq c$, then let $\left\{x_3,x_4,\ldots,x_n\right\}=[n]\setminus\{a,b,c\}$, where $x_3<x_4<\ldots<x_n$. We define $$f\big((a,b,c)\big):=\left(a,c,x_3,x_5,x_4,x_6,x_7,x_8,\ldots,x_n\right)\,.$$ If $a\neq b$ and $b=c$, then let $\left\{x_3,x_4,\ldots,x_n\right\}=[n]\setminus\{a,b,c\}$, where $x_3<x_4<\ldots<x_n$. We define $$f\big((a,b,c)\big):=\left(a,b,x_3,x_4,x_6,x_5,x_7,x_8,\ldots,x_n\right)\,.$$ If $a=c$ and $a\neq b$, then let $\left\{x_3,x_4,\ldots,x_n\right\}=[n]\setminus\{a,b,c\}$, where $x_3<x_4<\ldots<x_n$. We define $$f\big((a,b,c)\big):=\left(a,b,x_3,x_6,x_5,x_4,x_7,x_8,\ldots,x_n\right)\,.$$ Finally, if $a=b=c$, then let $\left\{x_2,x_3,\ldots,x_n\right\}=[n]\setminus\{a,b,c\}$, where $x_2<x_3<\ldots<x_n$. We define $$f\big((a,b,c)\big):=\left(a,x_2,x_3,x_6,x_4,x_5,x_7,x_8,\ldots,x_n\right)\,.$$ It is clear that $f$ is injective. Note that $f$ is not surjective because $$(1,2,3,5,6,4,7,8,9,\ldots,n)\notin\text{im}(f)\,.$$

P.S. The reason that this proof fails for $n\in\{1,2,3,4,5\}$ is that the definition of $f$ requires at least six coordinates.

Answer 4

An option (no induction).

Let $n \ge 6$, $n$ integer.

We want to show

$\frac{n(n-1)(n-2)}{n^3} (n-3)! > 1$;

$(1-1/n)(1-2/n)(n-3)! \ge$

$(1-1/6)(1-2/6)(n-3)! = (5/9)(n-3)! >$

$(1/2)(n-3)! >1$, since

$(n-3)! \ge 6$, for $n \ge 6$.