Question 69
Determine the maximum possible value of real number $k$, for which the inequality
$$
\frac{a}{1 + 9bc + k(b - c)^2} + \frac{b}{1 + 9ca + k(c - a)^2} + \frac{c}{1 + 9ab + k(a - b)^2} \ge \frac{1}{2}
$$
is satisfied for every choice of nonnegative real numbers $a, b, c$ satisfying $a + b + c = 1$.
$$
\frac{a}{1 + 9bc + k(b - c)^2} + \frac{b}{1 + 9ca + k(c - a)^2} + \frac{c}{1 + 9ab + k(a - b)^2} \ge \frac{1}{2}
$$
is satisfied for every choice of nonnegative real numbers $a, b, c$ satisfying $a + b + c = 1$.