This problem was supposed to be solved with proof by induction but I wanted to see if this was a valid way of doing it or if this method fails. I'm just starting my first proofs class so there is probably some silly mistakes. Is this solution valid?
Prove that $f(n)\lt g(n) \text{ when } n\ge3 \text{ and n is a apart of all natural numbers}$
$f(n)=2n^2+5\text{ and }g(n)=n^3$
$f(n)\lt g(n) \text{ when}$ $n\ge3$ if for $n=3$ ⟹ $f(n)\lt g(n)$ and $n\gt3$ ⟺ $f(n)\lt g(n)$
1.
$$n=3 \space; 2(3)^2+5\lt{(3)^3}$$
$$⟹23\lt27$$
2.
$$2n^2+5\lt{n^3}$$
$$⟹2n^2\lt{n^3}-5 \space\text{and } n\gt3\gt-5$$
$$⟹0\lt{n^3-2n^2-n}$$
$$⟹0\lt{n(n^2-2n-1)}$$
$$⟹0\lt{n^2-2n-1} \text{ and } n\gt0\text{ ; } n\gt3\gt-1$$
$$⟹0\lt{n^2-2n-n}$$
$$⟹0\lt{n(n-2-1)} \text{ and } n\gt0 \text{ ; }; n\gt3\gt-1$$
$$⟹0\lt{n-3}$$
$$⟹0\lt3\lt{n}$$
Therefore by 1 and 2 $f(n)<g(n)$ when $n≥3$
Induction proof: $f(n+1)=2(n+1)^2+5=f(n)+4n+2$ and $g(n+1)=(n+1)^3=g(n)+3n^2+3n+1$ For $n\gt 3$, $4n+2\lt 3n^2+3n+1$ plus induction assumption $f(n)\lt g(n)$ implies $f(n+1)\lt g(n+1)$.