By symmetry, WLOG $a\ge b\ge c$, by Cauchy-Bunyakovsky-Schwarz inequality. Consider \begin{align*}\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}&=\frac{\sqrt{(1-a)bc}+\sqrt{(1-b)ac}+\sqrt{(1-c)ab}}{\sqrt{abc}} \\&= \frac{\sqrt{(1-a)bc}+\sqrt{a}(\sqrt{(1-b)c}+\sqrt{b(1-c)})}{\sqrt{abc}} \\&\le\frac{\sqrt{(1-a)bc}+\sqrt{a}\sqrt{(1-b)+b}\sqrt{(1-c)+c}}{\sqrt{abc}} \\&= \frac{\sqrt{(1-a)bc}+\sqrt{a}}{\sqrt{abc}}\\&\le\frac{\sqrt{(1-a)+a}\sqrt{bc+1}}{\sqrt{abc}}\\&=\sqrt{\frac{bc+1}{abc}}\\&=\huge \min \Bigg\{ \sqrt{\frac{ab+1}{abc}}, \sqrt{\frac{bc+1}{abc}}, \sqrt{\frac{ac+1}{abc}} \Bigg \}\end{align*}So inequality holds when $\frac{1-b}{b}=\frac{c}{1-c}$ and $\frac{1-a}{a}=bc$ We get, $(a,b,c)=(\frac{1}{-t^2+t+1},t,1-t)$ for $\frac{1}{2}\le t<1$ and its permutation.

If I were them, I would stay quiet, to let such situation happen again. There are so many math contests, that repeating some task is obvious. I think this triple looks better $\left(\frac{t^2}{t^2+t-1},1-\frac{1}{t},\frac{1}{t}\right), t>1$ How did you get this $t\ge 0.5$, which in my version would be $t\le 2$?

WolfusA wrote: If I were them, I would stay quiet, to let such situation happen again. There are so many math contests, that repeating some task is obvious. Well, we actually plan to propose problems for the olympiad next year so it doesn’t happen.

By "we" you mean a committee or ex-contestants?

WolfusA wrote: By "we" you mean a committee or ex-contestants? We’re a bunch of ex-contestants compiling a shortlist of problems.