<s>[INST] Is this induction proof mathematically correct? Proof that $\frac{n^3}{3} < 3n-3$ is true for $n=2$ but false for every other $ n \in \mathbb{N}$.
Idea is to proof that $\frac{n^3}{3} \geq 3n-3$.
Let  $n$ be $n=3$
$\frac{27}{3}  \geq 9-3 $.
That means $\exists n \in \mathbb{N}$ such that $\frac{n^3}{3} \geq 3n-3$ and therefore the same for $n = n+1$.
We have to proof that $\frac{n^3}{3} >= 3n-3$ for $n=n+1$ which means $\frac{(n+1)^3}{3} \geq 3(n+1)-3 = \frac{(n+1)^3}{3} \geq 3n$
We have  $\frac{(n+1)^3}{3} = \frac{n^3+3n^2+3n+1}{3} = \frac{n^3}{3} + \frac{3n^2+3n+1}{3}$
Now we use that $\frac{n^3}{3} \geq 3n-3$.
$\frac{3n-3}{3} + \frac{3n^2+3n+1}{3} = \frac{3n^2+6n-2}{3} = n^2+2n-\frac23$
Can I now just say that $\forall n > 3; n \in \mathbb{N}$
$n^2+2n-\frac23 > 3n$
is true und therefore our statement $\frac{n^3}{3} < 3n-3$ is proven?
Thanks to everyone who takes some time to help me!
 [/INST] Dont write $n=n+1$ you can write $n\mapsto n+1$.
The typical way to estimate from $n^2+2n-\frac23$ is to use, that $n\geq 3$.
This helps dealing with the quadratic term. The rest is simple.
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