Team Selection Test for IMO

We prove that the minimum is $\frac{1}{3}$. Let $x_i>0$ for $i=1,2,3$ be the roots of the equation $x^3-a{x}^2+bx-c=0$. By Vieta's theorem $$x_1+x_2+x_3=a,x_1{x}_2+x_2{x}_3+x_1{x}_3=b,x_1{x}_2{x}_3=c.$$ Then $$A=\frac{1+a+b+c-c}{3+2a+b}=\frac{1}{b}=\frac{x_1{x}_2{x}_3}{x_1{x}_2+x_2{x}_3+x_1{x}_3}=\frac{1}{\frac{1}{x_1+1}+\frac{1}{x_2+1}+\frac{1}{x_3+1}}=\frac{\frac{1}{x_1(x_1+1)}+\frac{1}{x_2(x_2+1)}+\frac{1}{x_3(x_3+1)}}{\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right)\left(\frac{1}{x_1+1}+\frac{1}{x_2+1}+\frac{1}{x_3+1}\right)}$$ Without loss of generality we can assume that $0<x_1\le x_2\le x_3$. Then $$\frac{1}{x_1}\ge \frac{1}{x_2}\ge \frac{1}{x_3}>0\text{ and }\frac{1}{x_1+1}\ge \frac{1}{x_2+1}\ge \frac{1}{x_3+1}>0.\text{ By Chebyshev’s inequality}$$ Therefore, $A\ge 3$. The minimum $1/3$ of $A$ is reached for $x_1=x_2=x_3=t>0$ or $a=3t,b=3t^2,c=t^3$.