<s>[INST] Prove by induction $3^n\gt 5n^2$ for all $n \ge 4$ Prove by induction $3^n\gt 5n^2$ for all $n \ge 4$
Hello, I cannot seem to find this question, tried every search option (exact match), and not even wolframalpha could help me. Looking at a graph that is obviously true for all $n \ge 4$
My approach was trying to get a trinomial perfect square on the 5 side but it seems imposible with the tools of inequality.
 [/INST] Base case: $n=4$
$3^4 = 81 > 80 = 5\cdot 16 = 5 \cdot 4^2$.
For the inductive step, we'll first prove a lemma by induction as well: $2 \cdot 3^n > 10 n + 5, \forall n \geq 4$
Base case': $n=4$
$2 \cdot 3^4 = 2 \cdot 81 = 162 > 45 = 10 \cdot 4 + 5$.
Inductive step':
Suppose $2 \cdot 3^n > 10 n + 5$ holds for some $n \in \mathbb{N}^{\geq 4}$.
$2 \cdot 3^n > 10 n + 5 \implies 3(2 \cdot 3^n) > 3(10 n + 5) \implies 2 \cdot 3^{n+1} > 30 n + 15$
Since $n \geq 4 > 0$,
$2 \cdot 3^{n+1} > 10 n + 15 = 10 n + 10 + 5 = 10(n+1) + 5$
Therefore, we have proved the lemma.
Let's procede with the inductive step for our original statement:
Inductive step:
Suppose $3^n > 5 n^2$ holds for some $n \in \mathbb{N}^{\geq 4}$
$3^n > 5 n^2 \implies 3^n + 10 n + 5 > 5 n^2 + 10 n + 5 = 5 (n^2 + 2n + 1) = 5 (n+1)^2$
This implies that
$3^{n+1} = 3 \cdot 3^n = 3^n + 2 \cdot 3^n > 3^n + 10n + 5 > 5n^2 + 10n + 5 = 5 (n+1)^2$
We got $3^{n+1} > 5(n+1)^2$. Q.E.D.
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