Show that $\frac{3}{8}\le\int_0^{1/2}\sqrt{\frac{1-x}{1+x}}dx\le \frac{\sqrt{3}}{4}$ Show that $\displaystyle\frac{3}{8}\le\int_0^{1/2}\sqrt{\frac{1-x}{1+x}}dx\le \frac{\sqrt{3}}{4}$ without calculating the integral.
This is an exercise for training the use of mean value theorem for integral.
I tried to apply it in different ways, but the bounds I get are worst that those I need:


*

*A direct application does not work

*I also tried to bound first the function and then apply the thorem, but also failed.


I tried to split the interval and apply the Mean value theorem (in fact, is the same that take upper and lower Riemann sums).
Using MATHEMATICA, We need to split the interval in 3 parts (with equal length), to get (something better than) the upper bound; and we need to split the interval in 8 parts to get (something better than) the lower bound.
Now I conjecture there's some mistake/misprint in the original exercise.

We have:
$$\frac{d^2}{dx^2}\sqrt{\frac{1-x}{1+x}}=\frac{(1-2x)}{(1-x)^{3/2}(1+x)^{5/2}}$$
so the integrand function is convex over $\left[0,\frac{1}{2}\right]$. That implies (see Hermite-Hadamard inequality):
$$\frac{f\left(\frac{1}{4}\right)}{2}\leq\int_{0}^{\frac{1}{2}}f(x)\,dx \leq \frac{f(0)+f\left(\frac{1}{2}\right)}{4}\tag{1}$$
or:
$$\color{red}{0.387}\ldots=\frac{1}{2}\sqrt{\frac{3}{5}}\leq\color{red}{ \int_{0}^{\frac{1}{2}}\sqrt{\frac{1-x}{1+x}}\,dx} \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{3}}\right)=\color{red}{0.394}\ldots\tag{2}$$
that is way stronger that what was asked to prove.
