Mathematical Induction - Inequality

Question

Does anyone have any idea on how to complete the inductive step?

Thm: For all $n >= 0~~~~ 6^n + 4 > n^3$

Pf: by Induction
    Let $P(n)$ be proposition that $~6^n + 4 > n^3$

Base Case: $P(0)$ is true.
   $~~6^0 + 4 > 0^3$
   $~~5 > 0~~~$(proven!)

Inductive Step:
   For all $~~n >= 0$, show that $P(n) => P(n+1)$.
   Assume $P(n)$ is true for purposes of induction.
   (ie, assume $~~6^n + 4 > n^3$)
   Need to show that (6^(n+1))+4 $> (n+1)^3$

   $~~6^n + 4 > n^3$
   $~~6(6^n + 4) > 6n^3$
     (6^(n+1))+4(6) $>6n^3$
     (6^(n+1))+4 $>6n^3-24+4$
     (6^(n+1))+4 $>6n^3-20$ (I'm stuck here...)

This theorem is self conjectured because I want to prove this on my own without any solutions present yet it seems that I can't prove it. I have already thought of any possible axioms that could help me but still no luck. Any ideas, tips or solutions are highly appreciated.

Answer 1

We assume:

$6^n + 4 > n^3$

Thus, we want to prove:

$6^{n+1} + 4 > (n+1)^3$

From the hypothesis:

$6^n > n^3-4 \rightarrow 6^{n+1} > 6n^3 -24$

It suffices to show that: $6n^3-24 > (n+1)^3$

Expanding gives: $5n^3-3n^2-3n-21$

We want to show that this is greater than zero. However, I don't want to find the roots. Thus, we let $n=1$ and see: $5-3-3-21 < 0$ So $n=1$ does not work. However, letting $n=2$ gives $40-12-6-21 > 0$ Thus, we take the derivative of $5n^3-3n^2-3n-21$ and get: $15n^2-6n-3$ Since the parabola open up, and letting $n=2$ is positive, we can see that the derivitive is greater than 0 for $n>2$ and thus the original function is greater than zero for $n \ge 2$ Thus, we have proved it for $n \ge 2$ Substituting in $n=1, 0$ gives the complete solution set

Answer 2

All you have left to prove is that $6n^3-20 \geq (n+1)^3$

So it's sufficient to prove that $5n^3-3n^2-3n-21 \geq 0$

Now, simply investigate the function $5n^3-3n^2-3n-21$:

• Find the derivative of the function
• Show that the only root of the function is at $n<2$
• Show that the function is monotonously increasing for $n\geq2$
• Hence $\forall{n\geq2}:5n^3-3n^2-3n-21 \geq 0$, Hence $\forall{n\geq2}:6^n+4 > n^3$

Finally, you've already shown that $6^n+4>n^3$ for $n=0$, so just show it for $n=1$...