Question 81
Show that, if the numbers $a$, $b$, $c \in [0, \infty)$ satisfy the relation $a + b + c = 3$, then
$$
\frac{a}{3a + bc + 12} + \frac{b}{3b + ca + 12} + \frac{c}{3c + ab + 12} \le \frac{3}{16}.
$$
$$
\frac{a}{3a + bc + 12} + \frac{b}{3b + ca + 12} + \frac{c}{3c + ab + 12} \le \frac{3}{16}.
$$