Question 180
Let $a, b, c$ be three positive real numbers. Show that $$
\frac{a+b+3c}{3a+3b+2c} + \frac{a+3b+c}{3a+2b+3c} + \frac{3a+b+c}{2a+3b+3c} \geq \frac{15}{8}
$$
\frac{a+b+3c}{3a+3b+2c} + \frac{a+3b+c}{3a+2b+3c} + \frac{3a+b+c}{2a+3b+3c} \geq \frac{15}{8}
$$