Problem 1433

The left-hand side is symmetric with respect to $a$, $b$, $c$. Hence, we may assume that $a > b > c \ge 0$. Note that replacing $(a, b, c)$ with $(a-c, b-c, 0)$ lowers the value of the left-hand side, since the numerators of each of the fractions would decrease and the denominators remain the same. Therefore, to obtain the minimum possible value of the left-hand side, we may assume that $c=0$.
Then the left-hand side becomes
$$\frac{a^2}{b^2} + \frac{b^2}{a^2},$$
which yields, by the Arithmetic Mean - Geometric Mean Inequality,
$$\frac{a^2}{b^2} + \frac{b^2}{a^2} \ge 2\sqrt{\frac{a^2}{b^2} \cdot \frac{b^2}{a^2}} = 2,$$
with equality if and only if $a^2/b^2 = b^2/a^2$, or equivalently, $a^4 = b^4$. Since $a, b \ge 0$, $a = b$. But since no two of $a, b, c$ are equal, $a \ne b$. Hence, equality cannot hold. This yields
$$\frac{a^2}{b^2} + \frac{b^2}{a^2} > 2.$$
Ultimately, this implies the desired inequality. $\square$

$$\frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} - 2 = \frac{[a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b)]^2}{[(a-b)(b-c)(c-a)]^2}.$$
Then Schur's Inequality tells us that the numerator of the right-hand side cannot be zero. $\square$