<s>[INST] Prove $\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$ if $3\leq aI don't really know if I should use brute force or some kind of theorem, it comes on a calculus past exam and it says:
suppose: $3≤a<b≤8$ 
prove that $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$
 [/INST] Note: This answer was finished after farruhota's post and has a similar argument.

$\quad \frac{b-a}{6} \le \sqrt{1+b} - \sqrt{1+a} \; \text{ iff }$
$\quad\quad\quad 1 \le 6 \frac{\sqrt{1+b} - \sqrt{1+a}}{b-a} = \frac{6}{ \sqrt{1+b} + \sqrt{1+a}}$
Ok, the denominator, $\sqrt{1+b} + \sqrt{1+a}$, of the RHS of the last inequality is always less than
$$ \sqrt{1+8} + \sqrt{1+8} = 3 + 3$$
and $1 \le \frac{6}{3 + 3}$.
To show the second inequality holds:
$\quad \frac{b-a}{4} \ge \sqrt{1+b} - \sqrt{1+a} \; \text{ iff }$
$\quad\quad\quad 1 \ge 4 \frac{\sqrt{1+b} - \sqrt{1+a}}{b-a} = \frac{4}{ \sqrt{1+b} + \sqrt{1+a}}$
Ok, the denominator, $\sqrt{1+b} + \sqrt{1+a}$, of the RHS of the last inequality is always greater than than
$$ \sqrt{1+3} + \sqrt{1+3} = 2 + 2$$
and $1 \ge \frac{4}{2 + 2}$.

We've shown that the OP's (actually strict) inequalities hold.

</s>