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.
 A: You should approach this by breaking it down into each step.
Step 1: The base case.
You are trying to prove that it is true for $n \geq 4$, and so your base case should be for the lowest integer satisfying the inequality (4). This is as simple as direct computation of the two values. This should be true (it is since $81 > 80$).
Step 2: Assume that it is true for $n=k$. That is to say, assume $3^k > 5k^2$.
Step 3: $n = k + 1$
$\text{RHS} = 5(k+1)^2 = 5(k^2 + 2k + 1)$
As Manifoldski pointed out, $(n + 1)^2 < 3n^2$ for $n>2$
$\implies \text{RHS} < 3(5k^2)$
Since we have assumed step 2,
$\implies \text{RHS} < 3(5k^2) < 3(3^k) = 3^{k+1}$
$\implies \text{RHS} = 5(k+1)^2 < 3^{k+1} = \text{LHS}$
And thus the inequality holds for $n = k + 1$, given that it holds for $n = k$.
Since the inequality holds for $n = 4$, and if it holds for $n = k$, it also holds for $n = k + 1$, by induction, the inequality holds for all $n\in \mathbf{N},n\geq4$.
A: 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.
A: Alternative approach:
Part of the reason that I am posting this response is that this approach is not present in the MathSE posting linked to in the comment of José Carlos Santos, which immediately follows the original posting.

Suppose that you have two functions, $f(n)$ and $g(n)$ that are each defined on all positive integers.  Suppose further that you are given a constant $~N \in \Bbb{Z^+}~$ and are asked to prove that for all $~n \geq N, n \in \Bbb{Z^+},~$ that $f(n) > g(n)$.
After demonstrating that $~f(N) > g(N),~$ there are two generic approaches for completing the proof by induction.

*

*Show that for all $~n \in \Bbb{Z^+}, ~n \geq N,~$ you have that 
$f(n+1) - f(n) \geq g(n+1) - g(n).$


*Show that for all $~n \in \Bbb{Z^+}, ~n \geq N,~$ you have that 
$\displaystyle \frac{f(n+1)}{f(n)} \geq \frac{g(n+1)}{g(n)}.$
The second bullet pointed method above is an immediate result, for this particular problem.

So, setting $~f(n) = 3^n, ~g(n) = 5n^2,~$ the base case is:
$$3^4 = 81 > 80 = 5(4)^2.$$
Then,
$$\frac{f(n+1)}{f(n)} = 3$$
and, when $~n \geq 4$:
$$\frac{g(n+1)}{g(n)} = \left[\frac{(n+1)}{(n)}\right]^2 \leq \left[\frac{5}{4}\right]^2 < 3.$$
