Prove $\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$ if $3\leq a<b\leq 8$ [closed]

I 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: $$3≤a prove that $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$

closed as off-topic by Martin R, YuiTo Cheng, Alexander Gruber♦May 21 at 1:41

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• The usual approach is to expand with $\sqrt{1+b} + \sqrt{1+a}$ – Martin R May 20 at 14:58

Introduce:

$$1+a=x^2,\ \ 1+b=y^2$$

Obviously:

$$2\le x

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).

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 and we are done!

As Martin R suggested, multiply all sides by conjugate of the middle term (note: $$3≤a): $$\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\\$$

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.