<s>[INST] Prove that $3^{n + 1} \ge n^4 + n^2 + 1$ for any $n \ge 5$ Prove that $3^{n + 1} \ge n^4 + n^2 + 1$ for any $n \ge 5$, using mathematical induction.
I've thought about writing the equation like that: $3^{n + 1} \ge (n^2 + 1)^2 - n^2$, but all for nothing.
 [/INST] You can check the base case of $n=5$ - add that into your question.
For the inductive step, let's re-cast the inequality as $3^{n+1} - n^4-n^2-1 \geq 0$.
$$\begin{align}
3^{n+2} - (n+1)^4-(n+1)^2-1 &= \color{#0B2}{3^{n+1}} + 2\cdot3^{n+1} \color{#0B2}{- n^4} - 4n^3 - 6n^2 - 4n -1\\
&\quad\quad\quad \color{#0B2}{-n^2}-2n-1 \color{#0B2}{-1}\\
&\geq 2\cdot3^{n+1}  - 4n^3 - 6n^2 - 6n - 2\\
&= 2(3^{n+1} \color{#03B}{ - 2n^3 - 3n^2 - 3n - 1})\\
\end{align}$$
Now note that since $n>5$, $n^4+n^2+1 > 5n^3+5n+1 > \color{#03B}{2n^3+3n^2+3n+1}$ & finish
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