Prove by induction that for $n ≥ 3, 4^{n} > 5n^{2}+ n.$

The base case is clear to me, but I am not sure how to solve it for the inductive step

$n ≥ 3, 4^{n} > 5n^{2}+ n$

We must show that $ 4^{n+1} > 5(n+1)^{2}+ (n+1) $

We can start by multiplying by 4

$ 4 \times 4^{n} > 20n^{2}+ 4n$

Then we should show that

$ 20n^{2}+ 4n > 5(n+1)^{2}+ (n+1) = 5n^{2} + 6 + 11n$ and so that

$15k^{2} -7k- 6 > 0 $

But for this last inequality I can't manage to simplify properly in order to solve it. Can you give me any hint on how to proceed?

You can use the fact that$$15k^2-7k-6=0\iff k=\frac{7\pm\sqrt{409}}{30}$$and that therefore $15k^2-7k-6>0$ if $k>\frac{7+\sqrt{409}}{30}\approx0.91$. Actually, all you need is that$$\frac{7+\sqrt{409}}{30}<\frac{7+\sqrt{441}}{30}=\frac{7+21}{30}<1.$$

$$15k^{2}-7k-6=15[(k-\frac 7 {30})^{2}-\frac 6 {15}-\frac {49}{900}]$$ $$ \geq 15[(1-\frac 7 {30})^{2}-\frac 6 {15}-\frac {49}{900}]>0$$

It might be easier to argue as follows:

$$\begin{align} 5(n+1)^2+(n+1) &=5n^2+11n+6\\ &=(5n^2+n)+(10n+6)\\ &\le(5n^2+n)+(15n^2+3n)\quad(\text{since }10n\le15n^2\text{ and }6\le3n)\\ &=(5n^2+n)+3(5n^2+n)\\ &=4(5n^2+n)\\ &\lt4\cdot4^n\quad\text{(by induction)}\\ &=4^{n+1} \end{align}$$