Find $\max_{0 < \beta < 1/2} \{\min \{\alpha - \beta, 1 - 2\beta, \beta \alpha\}\}$ Hello I'm looking for find $$ \max\limits_{0 < \beta < 1/2} \{\min \{\alpha - \beta, 1 - 2\beta, \beta \alpha\}\} $$
where $\alpha \in (0,1)$ is fixed and $\alpha > \beta$.
I don't know how to handle with an inequality contrainst like $0 < \beta < 1/2 $.
 A: Two of them are decreasing with $\beta$, and the other is increasing.  So there are only five places to look: $\beta=0;\beta=1/2$; and where two of the formulas are equal.A: If $\beta=0$, the numbers are $\alpha,1,0$, and the minimum is $0$.
B: If $\beta=1/2$, the minimum is either $\alpha-1/2$ or $0$.
C: If $\alpha-\beta=1-2\beta$, then $\beta=1-\alpha$ (if $\alpha>1/2$) and the minimum is either $2\alpha-1$ or $\alpha-\alpha^2$.
D,E: etc.
There is a switchover point, which is the $\alpha$ for which the three lines (functions of $\beta$) intersect in one point.  For that $\alpha$, the two options in C, in D and in E are equal.
A: Since $\;0<\alpha<1\;,\;\;0<\beta<\frac12\;$ , you get
$$\begin{cases}-\frac12<\alpha-\beta<1\\{}\\0<1-2\beta<1\\{}\\0<\alpha\beta<\frac12\end{cases}$$
A: $f(\alpha)= max_{0<β<1/2}(min(α−β,1−2β,βα))$
$\alpha$ in $Open(0)$ .. $ \frac {\sqrt 5}{2}-\frac{1}{2})$ :
$[\alpha, \beta, f(\alpha)] = [\alpha, \frac{\alpha}{1+\alpha}, \frac{\alpha^2}{1+\alpha}]$
$\alpha$ in $\frac{\sqrt 5}{2}-\frac{1}{2}$ .. $Open(1)$ :
$[\alpha, \beta, f(\alpha)] = [\alpha, \frac{1}{2+\alpha}, \frac{\alpha}{2+\alpha}]$
