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 $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$

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! 
