Prove by mathematical induction that
$$\forall n \geq 4 \in \mathbb{N}: 3n^2 + 3n +1 < 2(3^n)$$
Step 1: Show that the statement is true for n = 4:
$$3(4)^2 + 3(4) +1 < 2(3^4)$$
Which simplifies to $61 < 162$. That completes the base case.
Step 2: Show that if the statement is true for $n = p$, it is true for $n = p + 1$:
$$3(p+1)^2 + 3(p+1) + 1 =$$ $$3(p^2 + 2p + 1) + 3p + 3 + 1 =$$ $$3p^2 + 3p + 1 + 6p + 6$$ $$ < 2(3^p) + 6p + 6$$
All that is required now is to show that $6p + 6 \leq 4(3^p)$ since $2(3^{p+1}) = 6(3^p)$. Does this require another proof by mathematical induction or is it sufficiently rigorous to say something like "well, $x^p$ clearly grows faster than $px$ as $p$ approaches infinity, so the inequality is trivial"?
In a larger context, are there any specific rules for when inequalities can be considered trivial?
By the inductive hypothesis, we assume that this statement is true:
$$3n^2 + 3n + 1 < 2(3^n)$$
$$3(n+1)^2 + 3(n+1) + 1 = 3n^2 + 9n + 7$$
From the inductive hypothesis:
$$3n^2 + 3n + 1 +6n+6 < 2(3^n)+6n+6$$ $$3(n+1)^2 + 3(n+1) + 1 < 2(3^n)+6n+6$$
So it will suffice to show:
$$2(3^n)+6n+6 < 2(3^{n+1}) = 6(3^n)$$ $$2(3^n)+6n+6 < 2(3^{n+1}) = 6(3^n)$$ $$6n+6 < 4(3^n)$$
Return to the inductive hypothesis:
$$6n^2 + 6n + 2 < 4(3^n)$$
As $6n^2$ is strictly increasing in our domain, and $6(4)^2 = 96 > 4$, we can replace it: $$4 + 6n + 2 < 6n^2+6n+2 < 4(3^n)$$ $$6n + 6 < 4(3^n)$$ And so we are done.