Prove $3^n$ > $n^4$ if $n\geq 8$

Question

As the title states, I am tasked with proving $3^n$ > $n^4$ if $n\geq 8$

The base case is trivial to prove. It is obvious that $3^8 > 8^4$ since $3^8 = 6561$ and $8^4 = 4096$, and $6561 > 4096$, thus the base case $n = 8$ is true.

Now for the Induction Hypothesis (IH) we let $n = m$ which then says $3^{m} > m^{4}$

I know that $3^{m+1} > (m+1)^{4}$ is what we are trying to show. So when you expand this you get $$3*3^{m} > m^{4} + 4m^{3} + 6m^{2} + 4m + 1$$ (Binomial Expansion Theorem is how you get the right hand side of the above inequality).

But this is where I'm stuck. I know that $3^{m} > m^{4}$ via the IH but idk what else to say or where to go. Any help would be appreciated. Thanks!

Answer 1

Note that ${m+1 \over m} = 1 + {1\over m}$ is decreasing as $m$ increases and is less than $\sqrt[4]{3} = 1.316...$

$3\cdot3^m > 3\cdot m^4 = (\sqrt[4]{3}\cdot m)^4 > (m+1)^4$

Answer 2

yes you have $$3n^4>n^4+4n^3+6n^2+4n+1$$ or $$2n^4-4n^3-6n^2-4n-1>0$$ and this is true for $$n\geq 8$$ the only real and positive Zero Point is $$n\approx 3.61381$$

Answer 3

Alternatively and perhaps easier to show and understand:

We first remind ourselves that $m\geq 8$

$(m+1)^4=m^4+\color{red}{4m^3}+6m^2+\color{blue}{4m+1}$

$<m^4+\color{red}{m^4}+6m^2+\color{blue}{m^2+m^2}=m^4+m^4+8m^2\leq m^4+m^4+m^4=3m^4$

Thus: $3^{m+1}=3\cdot 3^m>3\cdot m^4>(m+1)^4$

The big observations here are when $m\geq 8$ we have $4m^3<m^4$, that $4m<m^2$, and that $8m^2<m^4$. To see why each of these are true remember that $a<b$ implies that $ax<bx$ for all $x\gt 0$. Looking in more detail at $4m^3<m^4$ for example, we have $4\cdot m^3<8m^3\leq^{\color{grey}{\dagger}} m\cdot m^3=m^4$ where the inequality marked with the $\dagger$ is as a result of $8\leq m$

Answer 4

Pulling it together into a compact induction proof:

Base case: $3^8 = 9^4>8^4$

Inductive hypothesis: $3^k >k^4$ for some $k\ge 8$

$\begin{align} \text{Observe }(k+1)^4 &= k^4 + 4k^3 + 6k^2 + 4k + 1 \\ &< k^4 + (4+6+4+1)k^3 \\ &< k^4 + 16k^3 \\ & \le 3k^4 \end{align}$

Then $3^{k+1}=3\cdot3^k>3\cdot k^4> (k+1)^4$ as required