Let $f(x)=\dfrac{2}{a}\sqrt{1-x}\left(a^2+\dfrac{1}{1-2x}\right)\ \ 0\leq x <\dfrac{1}{2}$ such that $a>0$, then find the minimum of $f$.
My work:
$$f'(x)=-\dfrac{2\left(\left(2a^2\left(x-1\right)+1\right)x-2\right)+a^2+1}{a\sqrt{1-x}\left(2x-1\right)^2}$$
Now:
$$2\left(\left(2a^2\left(x-1\right)+1\right)x-2\right)+a^2+1=0$$
Then I found :
$$x=\dfrac{(4a^2-2)\pm \sqrt{(2-4a^2)^2-16a^2(a^2-3)}}{8a^2}$$
Now what do I do? Please help me.
Suppose $x_1$ is a root of $f'(x)=0$.
After that step, you should compare $0$, $x_1$, and $\frac 1 2$ to see that if $0<x_1<\frac 1 2$. Note that the value of $x_1$ is relies on $a$. If so, probably you will find that $f$ is decreasing on $(0,x_1)$ and increasing on $(x_1,\frac 1 2)$, as the derivative is negative and positive on these intervals respectively. Then, $f(x_1)$ would be the minimum value. If not, then $f$ should be monotonic on $(0,\frac 1 2)$. Since $\lim_{x\to\frac 1 2}f(x)=\infty$ and $f(\frac 1 2)$ does not exist, $f(0)$ is the minimum value.
How it looks like as $a=0.1, 0.2, \ldots 2$.
If you want to know the minimum number for every $a>0$, you should take the derivative of the minimum or find some other way to analyze.