Saudi Arabian Mathematical Competitions

First, we will show that $a$, $b$, $c$ are pairwise coprime. Denote $d = \operatorname{gcd}(b, c)$ and suppose that $d > 1$. Take $p$ as a prime divisor of $d$. We have
$$\left\{\begin{array}{l} p \mid b \\ p \mid c \end{array}\right. \Rightarrow \left\{\begin{array}{l} p \mid (c - a)^2 \\ p \mid (a - b)^2 \end{array}\right. \Rightarrow \left.\begin{array}{l} p \mid c - a \\ p \mid a - b \end{array} \Rightarrow p \mid a.\right.$$
This implies that $p \mid \operatorname{gcd}(a, b, c)$, a contradiction. So $\operatorname{gcd}(b, c) = 1$. Similar with $\operatorname{gcd}(c, a)$, $\operatorname{gcd}(a, b)$.

Suppose that $a$, $b$, $c$ are side-lengths of some triangle then
$$b + c > a, \quad c + a > b, \quad a + b > c.$$
So by putting $x = (a + b - c)(b + c - a)(c + a - b) > 0$, we can see that $x > 0$.

We have $a \mid (b - c)^2 - a^2 = (b - c - a)(b - c + a)$, thus $a \mid x$. Similarly, we have $b \mid x$, $c \mid x$ and since $\operatorname{gcd}(a, b) = \operatorname{gcd}(b, c) = \operatorname{gcd}(c, a) = 1$ so $a b c \mid x$. Hence,
$$a b c \leq (a + b - c)(b + c - a)(c + a - b).$$
From AM-GM, we have $(a + b - c)(b + c - a) \leq \left(\frac{a + b - c + b + c - a}{2}\right)^2 = b^2$ and similar inequalities as $(b + c - a)(c + a - b) \leq c^2$, $(c + a - b)(a + b - c) \leq a^2$ then
$$(a + b - c)(b + c - a)(c + a - b) \leq a b c.$$
These imply the equality must be hold, so $a = b = c$, contradiction since $a$, $b$, $c$ pairwise distinct. $\square$