Prove that $n! > n^{3}$ for every integer $n \ge 6$ using induction

Question

Proof.

Base case. When $n = 6$, LHS = $6! = 720$ and RHS = $6^{3} = 216$. Thus LHS > RHS.

Inductive step. Assume that $m! > m^{3}$ for some fixed integer $m$ such that $m \ge 6$. We must show that it also holds true for $(m+1)$, i.e, $(m+1)! > (m+1)^{3}$. Observe that

\begin{align*} (m+1)! &= (m+1) \cdot m! \\ &> (m+1) \cdot m^{3} \text{$\quad$(by the inductive hypothesis)}\\ \end{align*}

(still unfinished)

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I know it boils down to showing that $(m+1) \cdot m^{3} > (m+1)^{3}$ but I'm unsure on how to proceed.

Answer 1

If you want to show $(m+1)\cdot m^3>(m+1)^3$, this is equivalent to

$$m^3>(m+1)^2.$$

As $m>2$, this is implied by

$$2m^2>(m+1)^2,$$

which is equivalent to

$$m^2>2m+1$$

or

$$(m-1)^2>2,$$

which is true if $m\geq 3$.

(Note that if you were to write this up it would be clearer if you started from $(m-1)^2>2$, which you know is true, and worked back to $m^3>(m+1)^2$ -- this writeup serves to help make clear how one would come up with this.)