<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] As Martin R suggested, multiply all sides by conjugate of the middle term (note: $3≤a<b≤8$):
$$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4} \iff \\
\frac{\overbrace{\require{cancel}\cancel{b-a}}^{1}}{6}(\sqrt{1+b} + \sqrt{1+a})≤\overbrace{\cancel{b-a}}^{1}≤\frac{\overbrace{\cancel{b-a}}^{1}}{4}(\sqrt{1+b} + \sqrt{1+a})\iff \\
\text{LHS:} \ \frac{\sqrt{1+b} + \sqrt{1+a}}{6}<  \frac{\sqrt{1+b} + \sqrt{1+8}}{6}\le 1 \iff \sqrt{1+b}\le 3 \iff b\le 8 \ \ \color{red}\checkmark\\
\text{RHS:} \ \frac{\sqrt{1+b} + \sqrt{1+a}}{4}>\frac{\sqrt{1+3} + \sqrt{1+a}}{4}\ge 1 \iff \sqrt{1+a}\ge 2 \iff a\ge 3 \ \ \color{red}\checkmark\\
$$
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