Use math induction to show that
1) n!>= n^2 for n>=6

2) n!>=n^3 for n>=6

2. Originally Posted by xomichelleybelly
Use math induction to show that
1) n!>= n^2 for n>=6
I will assume that you have done the base case and that we assume this is true for some $\displaystyle k \ge 6$

So we need to show that:

$\displaystyle (k+1)! \ge (k+1)^2$,

well:

$\displaystyle (k+1)! = k!~ k$

and by supposition $\displaystyle k! \ge k^2$, so:

$\displaystyle (k+1)! = k! k \ge k^3$

So if we can prove that $\displaystyle k^3 \ge (k+1)^2$ we are done, but $\displaystyle k^3 \ge (k+1)^2$ is equvalent to $\displaystyle k^3-k^2-2k-1 \ge 0$. So to compete the proof of the induction step you need to show that:

$\displaystyle k^3-k^2-2k-1 = 0$

has no real root greater than or equal to $\displaystyle 6$ (this is sufficient as for $\displaystyle k$ greater than its largest real root the cubic is positive).

This you do by making the substitution $\displaystyle x=k-6$, then using Descartes rule of signs to show that this has no positive roots.

(In fact the only real root of $\displaystyle k^3-k^2-2k-1 = 0$ is near $\displaystyle k=2.1$)

RonL

3. Originally Posted by xomichelleybelly
Use math induction to show that

2) n!>=n^3 for n>=6
Exactly the same method as for part 1, but the equation that you need to
show has no root greater than 6 is:

$\displaystyle k^4-k^2-2k-1=0$

and exactly the same method will work with this.

RonL

4. Originally Posted by xomichelleybelly
Use math induction to show that
1) n!>= n^2 for n>=6

2) n!>=n^3 for n>=6

By proving 2) you've automatically proved 1).

(And I reckon a lawyer could argue that induction has been used to prove both )