Question 147
Let positive numbers $a$, $b$, $c$ be given. Prove that
$$
\frac{a}{3a^2 + b^2 + 2ac} + \frac{b}{3b^2 + c^2 + 2ab} + \frac{c}{3c^2 + a^2 + 2bc} \le \frac{3}{2(a+b+c)}
$$
$$
\frac{a}{3a^2 + b^2 + 2ac} + \frac{b}{3b^2 + c^2 + 2ab} + \frac{c}{3c^2 + a^2 + 2bc} \le \frac{3}{2(a+b+c)}
$$