Question 107
Let $a$, $b$ and $c$ be positive real numbers with $c \ge b \ge a$. Prove that
$$
\frac{(c-a)^2}{6c} \le \frac{a+b+c}{3} - \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}
$$
$$
\frac{(c-a)^2}{6c} \le \frac{a+b+c}{3} - \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}
$$