Question 9
Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 2$. Prove that
$$
\frac{(a-1)^2}{b} + \frac{(b-1)^2}{c} + \frac{(c-1)^2}{a} \ge \frac{1}{4} \left( \frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c} + \frac{c^2+a^2}{c+a} \right).
$$
$$
\frac{(a-1)^2}{b} + \frac{(b-1)^2}{c} + \frac{(c-1)^2}{a} \ge \frac{1}{4} \left( \frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c} + \frac{c^2+a^2}{c+a} \right).
$$