Question 9
Let $a$, $b$, $c$ be real numbers for which $a + b + c = 4$ and $a, b, c \gt 1$. Prove that
$$
\frac{1}{a-1} + \frac{1}{b-1} + \frac{1}{c-1} \ge 8 \left( \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \right).
$$
Macedonian source version, translated in full:
Let $a$, $b$, $c$ be real numbers for which $a + b + c = 4$ and $a, b, c \gt 1$. Prove that
$$
\frac{1}{a-1} + \frac{1}{b-1} + \frac{1}{c-1} \ge 8 \left( \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \right).
$$
$$
\frac{1}{a-1} + \frac{1}{b-1} + \frac{1}{c-1} \ge 8 \left( \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \right).
$$
Macedonian source version, translated in full:
Let $a$, $b$, $c$ be real numbers for which $a + b + c = 4$ and $a, b, c \gt 1$. Prove that
$$
\frac{1}{a-1} + \frac{1}{b-1} + \frac{1}{c-1} \ge 8 \left( \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \right).
$$