I am trying to prove the inequality in the title for $n\geq 4$; however, I am stuck on the induction step! Any help would be appreciated.
For $n\ge 4$, prove that $n! > n^2$.
Base Case: $n=4$, LHS $4! = 24$, RHS = $4^2 = 16$
$24>16$ : True
Induction Hypothesis: Assume True for $n=k$.
$k! > k^2$
Induction Step: Should be True for $n=k+1$
$(k+1)! > (k+1)^2$
$(k+1) . (k)! > (k^2 + 2k + 1)$
However, here is where I get stuck.
Hint:
$$(k+1)!> (k+1)^2$$ can be rewritten as
$$(k+1)\cdot (k!) > (k+1)\cdot(k+1)$$
For the induction step, here is the key part: \begin{align} (k+1)! &= (k+1)\cdot k!\tag{by definition}\\[0.5em] &> (k+1)\cdot k^2\tag{by the inductive hypothesis}\\[0.5em] &= k^3+k^2\tag{expand}\\[0.5em] &> k^2+2k+1\tag{since $k\geq 4$}\\[0.5em] &= (k+1)^2. \end{align} The main problem is realizing that you can deduce that $k^3+k^2>k^2+2k+1$ from the fact that $k\geq 4$.
Assume, for $n = k$, the inequality above holds.
It will be shown, that for $n = k + 1$, the inequality above also holds.
Since $$k! > k^2,$$ then $$(k + 1)k! > (k + 1)k^2,$$ which can be rewritten as $$(k + 1)! > (k^3 + k^2)$$
Since $k \ge 4$, $k^2 \ge 16$, and it follows that $k^3 \ge 16k$.
Additionally, $14k > 1$, and $k^3 > k^2$
So, $$k^3+k^2 \ge k^3 + 16k = k^3 + 2k + 14k > k^2 + 2k + 1$$ Hence, $$(k + 1)! > (k^3 + k^2) > k^2 + 2k + 1 = (k +1)^2 $$.
$$(k+1)k^2 > (k+1)^2$$
That is
$$k^2 > k + 1$$
– Simon S Apr 27 '15 at 12:15