Prove by induction that $3^{n-2} \geq n^5$ for $n\geq20$ I am new to induction proofs and wanted to know if my reasoning behind the following proof is correct. Here are my steps :

*

*$n=18 \Rightarrow 3^{18} \geq 20^5$

*Given that $3^{n-2} \geq n^5$ is true for n, show $3^{n-1} \geq (n+1)^5$
$3^{n-1} = 3\times(3)^{n-2} \geq 3n^5$
If $3n^5 \geq (n+1)^5$, then $3\times (3)^{n-2} \geq 3n^5 \geq (n+1)^5 \Leftrightarrow 3^{n-1} \geq (n+1)^5$
I show this by induction for all $n \geq 20$.

*

*$n=18 \Rightarrow 3(20)^{5} \geq 21^5$


*$3(n+1)^{5} = 3n^5+15n^4+30n^3+30n^2+15n+3 \geq (n+1)^5+15n^4+30n^3+30n^2+15n+3$
$3n^5+15n^4+30n^3+30n^2+15n+3 \geq (n+2)^5-10n^4-40n^2-60n-28$
$(3n^5+25n^4+70n^3+30n^2+75n+31) = (n+2)^5 + (2n^5+15n^4+30n^3+10n^2+10n+1)\geq (n+2)^5$
Since $(2n^5+15n^4+30n^3+10n^2+10n+1) > 0$ for $n\geq20$, it follows that $(n+2)^5+(2n^5+...+1) \geq (n+2)^5$.
Therefore, $3\times (3)^{n-2} \geq 3n^5 \geq (n+1)^5 \Leftrightarrow 3^{n-1} \geq (n+1)^5$
 A: It might be easier to bypass the second induction and instead show directly that, for $n\ge20$, we have
$$\begin{align}
(n+1)^5&=n^5+5n^4+10n^3+10n^2+5n+1\\
&\le n^5+40n^4\\
&=n^5+2(20n^4)\\
&\le n^5+2(n\cdot n^4)\\
&=3n^5
\end{align}$$
where the first inequality (introducing the $40n^4$) holds for $n\ge1$ while the second inequality invokes the assumption $n\ge20$.
A: It seems like the limit is less ambitious than it could be. Let's aim for $n=15$.
Here is a $n=15$ base case to show $15^5 < 3^{13} $:
$$\begin{align}
15^5 &= 3^5\cdot 5^5\\
&<3^{5}\cdot 27^2\cdot 5\\
&=3^{11}\cdot 5\\
&<3^{13}\\
\end{align}$$
Then for $n\ge 15$,
$$\begin{align}
(n+1)^5 &= n^5 + 5n^4 + 10n^3+10n^2+5n+1 \\
 &< n^5 + 5n^4 + 10n^3+10n^2+6n \\
 &< n^5 + 5n^4 + 10n^3+11n^2 \\
 &< n^5 + 5n^4 + 11n^3 \\
 &< n^5 + 6n^4 \\
\color{#0B2}{\small(\text{so far only needing }n>11)}\qquad\qquad &< 2n^5  \\
\color{#0B2}{\small\text{inductive hypothesis}}\qquad \qquad  &< 2\cdot 3^{n-2}  \\
 &< 3^{n-1}
\end{align}$$
as required.
