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 $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$
 A: Introduce:
$$1+a=x^2,\ \ 1+b=y^2$$ 
Obviously:
$$2\le x<y\le3$$
Notice that:
$$4\le y + x \le 6\tag{1}$$
Inequality now becomes:
$$\frac{y^2-x^2}{6}\le y-x\le\frac{y^2-x^2}{4}$$
$$\frac{y+x}{6}\le 1\le\frac{y+x}{4}$$
...which is true because of (1).
A: By the mean value theorem and since $(\sqrt{1+x})'=\frac{1}{2\sqrt{1+x}},$ we obtain: $$\frac{\sqrt{1+b}-\sqrt{1+a}}{b-a}=\frac{1}{2\sqrt{1+c}},$$ where $3<c<8$ and we are done! 
A: As Martin R suggested, multiply all sides by conjugate of the middle term (note: $3≤a<b≤8$):
$$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4} \iff \\
\frac{\overbrace{\require{cancel}\cancel{b-a}}^{1}}{6}(\sqrt{1+b} + \sqrt{1+a})≤\overbrace{\cancel{b-a}}^{1}≤\frac{\overbrace{\cancel{b-a}}^{1}}{4}(\sqrt{1+b} + \sqrt{1+a})\iff \\
\text{LHS:} \ \frac{\sqrt{1+b} + \sqrt{1+a}}{6}<  \frac{\sqrt{1+b} + \sqrt{1+8}}{6}\le 1 \iff \sqrt{1+b}\le 3 \iff b\le 8 \ \ \color{red}\checkmark\\
\text{RHS:} \ \frac{\sqrt{1+b} + \sqrt{1+a}}{4}>\frac{\sqrt{1+3} + \sqrt{1+a}}{4}\ge 1 \iff \sqrt{1+a}\ge 2 \iff a\ge 3 \ \ \color{red}\checkmark\\
$$
A: 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.

