So I'm doing homework for my logic class... For what values of n element of natural numbers is $3^{n+1}>n^4$ and we are supposed to use induction. This is what I have so far...
Proof: The inequality holds for n=1. Since $3^2>1^4= 9>1$. Assume $3^{k+1}>k$, where $k\geq 1$. We show $3^{k+2}>(k+1)^4$
$3^{k+2}=3^2+3^k>3k^4=k^4+k^4+k^4...$ I'm not sure what to do from here.
The inequality $3^{n+1}>n^4$ is true for $n=1,\ 2$ and all $n\ge5$. (The inequality is false for $n=3$ and $n=4$ because $81\not>81$ and $243\not>256$.)
We can prove the inequality for $n\ge5$ by induction.
The base case ($n=5$) is true: $3^6>5^4, \ \ 729>625$.
Now suppose the inequality is true for some $k\ge5$: $$ 3^{k+1} > k^4. \tag{1} $$ But $$ 3^{k+2}=3^{k+1}+2\cdot3^{k+1}, \quad\mbox{ while }\quad (k+1)^4 = k^4 + 4k^3+ 6k^2+ 4k+ 1, $$ and for every $k\ge5$ we have $$ 2\cdot3^{k+1} > 2\cdot k^4 \ge 2\cdot5k^3> 4k^3+ 6k^2+ 4k+ 1 $$ by hypothesis $(1)$. Thus we have $3^{k+2}>(k+1)^4$; our induction step is completed.
Therefore, by the principle of mathematical induction, our inequality $$3^{n+1}>n^4$$ is true for all $n\ge5$.