Show $∀n≥3$, $2n^2+1 ≥ 5n$

Question

I was able to prove the base case statement, where if you plug in $3$ for $n$ you get:

$19 ≥ 15$.

Next I supposed an arbitrary value $k$ where $k ≥ 3$ and $2k^2+1 ≥ 5k$. I know that next I need to prove that:

$2(k+1)^2+1 ≥ 5(k+1)$

But this is where I got stuck.

Answer 1

$$2(n+1)^2+1=2n^2+4n+2+1\ge 5n+4n+2=9n+2$$

We are done, if we have $9n+2\ge 5n+5$ , which is true for $n\ge \frac{3}{4}$, so it is also true for $n\ge 3$. This completes the proof.

Even better is to argue $5n+4n+2\ge 5n+5$ because of $4n+2\ge 5$

Answer 2

Method I: Solving the quadratic $f(x)=2x^2-5x+1=0$ yields two solutions, approximately $.22\;\&\;2.28$ Thus for $x>2.28$ we have $f(x)>0$.

Method II: (induction). We let $f(n)=2n^2-5n+1$ and observe that $f(3)>0$. Now suppose that $f(n)>0$, we wish to verify that $f(n+1)>0$ but $$f(n+1)=2(n+1)^2-5(n+1)+1=f(n)+4n-3$$

As $f(n)>0$ (by the induction hypothesis) and $n>3\implies 4n-3>0$ we see that $f(n+1)>0$ as desired.