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

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).
