CMO 2017

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. $\square$