# Prove by induction. Show that if $n ≥ 6$, then $n! > n^3$

Show that if $$n ≥ 6$$, then $$n! > n^3$$

Initial Step: $$n = 6$$

LHS: $$6!=720$$ RHS: $$6^3=216$$ LHS > RHS

Inductive Step: Assume $$n=k$$ is true

$$k! > k^3$$

Prove $$n=k+1$$ is true

$$(k+1)! > (k+1)^3$$

Can you help me? I don't know where to go from here. I'm stuck in here.

• Can you prove $(7/6)^3<2\le k+1$ for $k\ge6$? What does that tell you about $((k+1)/k)^3$? – J.G. Oct 6 '20 at 16:53

Assume that $$k! > k^3$$ ...$$(1)$$

Let's show $$(k+1)k! > (k+1)^3$$

Multiply $$(1)$$ by $$(k+1)$$

$$(k+1)k! > (k+1)k^3$$

$$(k+1)! > (k+1)k^3>(k+1)(k+1)^2=(k+1)^3$$

Using

$$k^3>(k+1)^2$$

$$k^3-k^2+2k-1>0$$, $$k^3>k^2$$, and, $$2k-1>0$$ given that $$k\geq6$$

It suffices to show that $$k^3>(k+1)^2$$ for $$k≥6$$ and you can do that using induction.

Base case of $$k=6$$ proved.

Moving forward, LHS increases by a factor of $$(k+1).$$

RHS increases by a factor of $$\left(\frac{k+1}{k}\right)^3.$$

Although it can be argued that $$k=7$$ and $$k=8$$ must be independently verified, thereafter, with $$\frac{k+1}{k} < 2$$,

RHS increasing by smaller factor than LHS.

\begin{align} (k+1!) &> (k+1)^3 \\ k! (k+1) &> (k+1)^3 \\ k! &> (k+1)^2 \end{align}

Our inductive hypotesis is $$k! > k^3$$.

$$k^3 > (k+1)^2$$ $$k^3 - (k+1)^2 > 0$$ (1)

$$2.148$$ is a root of $$f(k) = k^3-(k+1)^2$$. Dividing $$f(k)$$ by $$(x - 2.148)$$ you can see that the discriminant of the second order polinomial is negative, so it's the only root. Also, it's derivarive $$f'(k) = 3k^2 - 2(k+1)$$ is positive for $$k> 1.215$$.

So, (1) is true for $$k> 2.148$$. In particular, for $$k>6$$.