Problem 2215

We prove that the minimum is $\frac{1}{3}$.
Let $x_i > 0$ for $i = 1, 2, 3$ be the roots of the equation $x^3 - a x^2 + b x - c = 0$. By Vieta's theorem
$$x_1 + x_2 + x_3 = a, \quad x_1 x_2 + x_2 x_3 + x_1 x_3 = b, \quad x_1 x_2 x_3 = c.$$
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 $a = 3t, b = 3t^2, c = t^3$.