Show the next inequality

Question

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? $$\frac{|a+b|-|a-b|}{ab}=\frac{2}{\max\{|a|,|b|\}}$$

Answer 1

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 $\left|\frac{a}{b}\right| \le 1$?

Answer 2

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 $.