<s>[INST] Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing. 
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.

I do not have use derivatives, so I decided to apply the definition of being a strictly increasing function, which should be: 

If we pick 2 numbers $a$ and $b$ from the domain of a function $f$, where $a < b$, then $f(a) < f(b)$.

Now (if that is a correct definition), I have tried to apply it to my case:

Let $a$, $b \in [-1, 1]$ and $a < b$. We want to show that $f(a) < f(b)$ or that $f(b) - f(a) > 0$.
We know that $x^2 + 3x + 2 = (x + 2)(x + 1)$, thus we have that $f(a) = (a + 2)(a + 1)$ and $f(b) = (b + 2)(b + 1)$ , therefore we need to show that:
$$(b + 2)(b + 1) - (a + 2)(a + 1) > 0$$
We can see that $(b + 2)$ and $(a + 2)$ will always be positive, and that $(b + 2) > (a + 2)$ , since $b > a$ (by assumption).
Since $b > a$, we know that $b > -1$ (otherwise $a \geq b$), thus $(b + 1) > 0$ (so we know that $(b + 2)(b + 1) > 0$ .  We also have at most $(a + 1) = 0$, and thus  we have that $(a + 2)(a + 1) \ge 0$.
So, here is the proof that  $f(b) - f(a) > 0$ .  

Am I correct? If yes, what can I improve it? If not, where are the erros and possible solutions?
 [/INST] Seems good to me. I would be lazier, though. Pick $a,b\in[-1,1]$, with $a<b$. Since $a \geq -1$, then $a+1 \geq 0$. Since $b > a$, we have also that $b+2,b+1,a+2 > 0$. So: $$\begin{cases} b+2>a+2 >0 \\ b+1 > a+1 \geq 0 \end{cases} \implies (b+2)(b+1) > (a+2)(a+1) \implies f(b) > f(a).$$
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