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?
 A: It looks right. Another possible approach is the following one - since:
$$ f(x)=x^2+3x+2 = \left(x+\frac{3}{2}\right)^2-\frac{1}{4} $$
we have that $f(x)$ attains its minimum in $x=-\frac{3}{2}$ (the abscissa of the vertex) and for every $r>-\frac{1}{4}$ the equation
$$ f(x)=r $$
has two solutions, symmetric with respect to $x=-\frac{3}{2}$. 
It follows that $f(x)$ is increasing over $\left(-\frac{3}{2},+\infty\right)$ and $[-1,1]$ is just a subset of it.

Still another way - for every $\varepsilon>0$ we have:
$$ f(x+\varepsilon)-f(x) = \varepsilon(2x+\varepsilon)+3\varepsilon>\varepsilon(2x+3) $$
hence $x\in[-1,1]$ implies $(2x+3)>0$, then $f(x+\varepsilon)>f(x)$, i.e. $f$ increasing.
A: The proof given looks fine to me, modulo my comment (see above).
Here's another way to see it, sans calculus:
Choose $x_1, x_2 \in [-1, 1]$ with $x_1 > x_2$.  Then
$f(x_1) - f(x_2) = x_1^2 + 3x_1 + 2 - x_2^2 - 3x_2 - 2 = x_1^2 - x_2^2 + 3(x_1 - x_2)$
$= (x_1 + x_2)(x_1 - x_2) + 3(x_1 - x_2) = (x_1 + x_2 + 3)(x_1 - x_2); \tag{1}$
note that $x_1 - x_2 > 0$ since $x_1 > x_2$ and that $x_1 + x_2 + 3 > 0$ since $x_1 + x_2 > -2$ since $x_1 > x_2 \ge -1$.  Thus
$f(x_1) -f(x_2) > 0, \tag{2}$
or
$f(x_1) > f(x_2), \tag{3}$
and we're done!
Note that all we really needed was $x_1 > x_2 \ge -1$.
Of course, with calculus we see that $f'(x) = 2x + 3 >0$ for $x \ge -1$, etc. etc. etc.  And since the minimum occurs at $x = -3/2$, Mark Bennet's comment is corroborated.
A: You are quite confused in your answer, though your edit has improved it. But it is still unclear how you are using the range of values you have been given.
What you should try to do, for my preference, is to reduce your expression to something which is obviously positive. As it happens, this is not difficult. We have $$(b^2+3b+2)-(a^2+3a+2)=(b^2-a^2)+3(b-a)=(b-a)(b+a+3)$$
If you are dealing with polynomials you will always find that $b-a$ factor arising in the difference, and this will always be useful in determining where the function is increasing or decreasing, since you can control the sign by choosing $b\gt a$ and you have reduced the degree of the expression you need to consider - here it takes you from quadratic to linear, which is easy.
A: Yes, to see more clearly, let $x=a+2, y =a+1, x+c = b+2, y+c = c+2$, compute $(x+c)(y+c)- xy$. Notice that $x, c$ are positive, $y$ is non-negative.
A: 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: $$\begin{cases} b+2>a+2 >0 \\ b+1 > a+1 \geq 0 \end{cases} \implies (b+2)(b+1) > (a+2)(a+1) \implies f(b) > f(a).$$
