Question 168
Prove that for all positive real numbers $a$, $b$, $c$ the inequality
$$
\frac{9b+4c}{11a^2} + \frac{9c+4a}{11b^2} + \frac{9a+4b}{11c^2} \ge \frac{1}{a} + \frac{1}{b} + \frac{1}{c}
$$
holds.
$$
\frac{9b+4c}{11a^2} + \frac{9c+4a}{11b^2} + \frac{9a+4b}{11c^2} \ge \frac{1}{a} + \frac{1}{b} + \frac{1}{c}
$$
holds.