Show the next inequality Show that:
$$
\frac{3}{8}\leq \int_{0}^{1/2}\sqrt{\frac{1-x}{1+x}}dx\leq \frac{\sqrt{3}}{4}.
$$
I have
$$m(b-a)\leq\int_{a}^{b}f(x)dx\leq M(b-a),$$
but the function does not present critical points in its derivative.
And how can I show the following?
$$\frac{|a+b|-|a-b|}{ab}=\frac{2}{\max\{|a|,|b|\}}$$
 A: For the first part, there seems to be some typing mistake. 
For the second, presume that $\left|\frac{b}{a}\right| \le 1$. Then
$$\frac{|a+b|-|a-b|}{ab}=\frac{|a||1+b/a|-|a||1-b/a|}{ab} = |a|\frac{1+b/a-1+b/a}{ab}=\frac{2|a|}{a^2}=\frac{2}{|a|}.$$
Can you figure out what happens when $\left|\frac{a}{b}\right| \le 1$?
A: If you have
written it correctly,
$\frac{3}{8}
\leq \int_{0}^{1/2}\sqrt{\frac{1-x}{1+2}}dx
\leq \frac{\sqrt{3}}{4}
$
is the same as
$\frac{3}{8}\sqrt{3}
\leq \int_{0}^{1/2}\sqrt{1-x}dx
\leq \frac{\sqrt{3}}{4}\sqrt{3}
$
or
$\frac{3\sqrt{3}}{8}
\leq \int_{0}^{1/2}\sqrt{1-x}dx
\leq \frac{3}{4}
$.
$\sqrt{1-x}$
is monotonic decreasing,
so,
on $[0, \frac12]$,
$1 \ge \sqrt{1-x} \ge \frac{\sqrt{2}}{2}$.
Therefore
$\frac1{2\sqrt{3}}
\approx 0.2886
\geq \int_{0}^{1/2}\sqrt{\frac{1-x}{1+2}}dx
\geq \frac{\sqrt{2}}{4\sqrt{3}}
\approx 0.20412
$.
If the denominator
is actually $x+2$,
then
$\sqrt{\frac{1-x}{x+2}}
$
is again monotonic decreasing
so,
on $[0, \frac12]$,
$\sqrt{\frac12}
\ge\sqrt{\frac{1-x}{x+2}}
\ge \sqrt{\frac{\frac12}{\frac52}}
= \sqrt{\frac15}
$
so the integral
is within half of these bounds.
For the second question,
use
$\min(a, b)
=\frac12(|a+b|-|a-b|)
$
and
$\max(a, b)\min(a, b)
=ab
$.
