Prove that for all natural numbers $n \geq 2$, $n^2 \geq n + 2$ Prove that for all natural numbers $n \geq 2$, $n^2 \geq n + 2$.
In solving this proof, I tried two methods, first working backward and then using the contradiction approach. i wish to solve this proof directly or using a contradiction however I am stuck on how to verify it for $n\geq 2$.
 A: for $n\geq2$, $n-2\geq 0$ and $n+1 > 0$
then
$(n-2)(n+1) \geq 0$
$n^2-n-2 \geq 0$
$n^2 \geq n+2$
A: For
$n \ge 2, \tag 1$
$n^2 = nn \ge 2n = n + n \ge n + 2. \tag 2$
Generalization:
If
$n \ge m \ge 2, \tag 3$
then
$n^2 = nn \ge mn = \displaystyle \sum_1^m n = n + \sum_1^{m - 1} n \ge n + m. \tag 4$
A: Another way to solve for the sake of curiosity. For $n\geq 2$, one gets:
\begin{align*}
\frac{n(n - 1)}{2} = 1 + 2 + 3 + \ldots + n- 1 \geq 1 \Rightarrow \frac{n(n- 1)}{2} \geq 1 \Rightarrow n^{2} - n \geq 2 \Rightarrow n^{2} \geq n + 2
\end{align*}
Hopefully this helps!
A: complete the square.
$n^2 - n \ge 2 \implies (n-\frac{1}{2})^2 \ge \frac{9}{4} \implies n \ge 2$ which is true.
A: Perhaps the quickest way:
If $n \ge 2$ then $n^2 = n \times n \ge n\times 2 = n + n \ge n + 2$.
If that was too slick:
We now that $n^2 > n$ in general.  But by how much and when?  $n^2 - n = n(n-1)$.  As $n \ge 2$ and $n-1 \ge 1$ then $n^2 - n \ge 2\times 1$ and $n^2 \ge n + 2$.
A cute thing happens when we try a proof by contradiction.
Suppose $n^2 < n + 2$ and $n\ge 2$.  If we divide both sides by $n > 0$ we get $n < 1 + \frac 2n$.  But as $n \ge 2$ we have $\frac 2n \le 1$ so $n < 1 + \frac 2n \le 1 + 1 =2$ so $n < 2$ which is a contradiction.
A creative way:  $n^2 = n^2$ and $n^2 - 1 = (n+1)(n-1)$.  And $n-1 \ge 1$ we have $n^2 -1 = (n+1)(n-1) \ge (n+1)\times 1 = (n+1)$ so $n^2 = n + 2$.
And there's always figuring out necessary and sufficient and/or equivalent statements.
$n^2 \ge n + 2 \iff n^2 - n \ge 2 \iff n(n-1) \ge 2$ and is that true?  As $n \ge 2$ and $n-1\ge 1$ it is.
Then there is always induction:
If $n = 2$ then $n^2 =4$ and $n + 2 = 4$ so $n^2 \ge n+2$.
If we assume that $n^2 \ge n+2$ then we have $(n+1)^2 = n^2 + 2n + 1$ and as $n^2 \ge n+2$ we have $(n+1)^2 = n^2 + 2n + 1 \ge (n+2) + 2n+ 1 = 3n + 3$ but is $3n +3 \ge 2(n+1) + 2$?  Well,  $3n + 3 = 2(n+1) + (n+1)$ and as $n \ge 2$ we have $2(n+1) +(n+1)\ge 2(n+1) + 3 > 2(n+1) + 2$.
[i.e.  $(n+1)^2 = n^2 + 2n + 1 \ge (2n+2) + (2n +1)=2(n+1) + (2n+1)> 2(n+1) + 2$]
So our induction step holds.
