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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: $3≤a<b≤8$ prove that $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$

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

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

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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!

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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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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.

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