Prove using induction: $n^2 - 7n + 12 \geq 0$, where $n \geq 3$ Does this work as an inductive prood? It feels like of weird, I might be doing the basis step twice unecessarily... 


If for every integer $n \geq 3$ we let $P(n)$ be the statement $n^2 - 7n + 12 \geq 0$, then to prove that $P(n)$ is true for all such $n$, we need to show that $P(3)$ and $\forall n \geq 3(P(n) \implies P(n+1))$ are true. In any proof by mathematical induction, a person doesn't need to carry around the notation "$P(n)$". In fact, doing so makes the proof ugly, so let's not include it below.
Proof.
If $n=3$, then $n^2-7n+12 = 9 - 21+12 = 0 \geq 0$.
Now let $n$ be an arbitrary integer satisfying $n \geq 3$, and suppose that $n^2 - 7n + 12 \geq 0$. Then
$$\begin{align*} 
(n+1)^2 - 7(n+1)+12 
&= (n^2+2n+1)-(7n+7)+12 &&\\ 
&= (n^2 - 7n + 12) + (2n + 1 - 7) &&\text{(after rearranging)}\\
&\geq 2n + 1 - 7 &&\text{(since $n^2 - 7n + 12 \geq 0$)}\\
&\geq 2(3)+1-7 &&\text{(since $n \geq 3$)}\\
&= 0.&&
\end{align*}$$
Therefore, by mathematical induction, $n^2 - 7n + 12 \geq 0$ is true for every integer $n \geq 3$.
