Question 31
For arbitrary positive numbers $a$, $b$, $c$ prove the inequality
$$
\frac{ab}{a+3b+2c} + \frac{bc}{b+3c+2a} + \frac{ca}{c+3a+2b} \le \frac{1}{6}(a+b+c).
$$
$$
\frac{ab}{a+3b+2c} + \frac{bc}{b+3c+2a} + \frac{ca}{c+3a+2b} \le \frac{1}{6}(a+b+c).
$$