Without loss of generality, let's assume $a$ is the the smallest among these three numbers. Then there must exist two positive real number $x$ and $y$ so that $b=a+x, c=a+y$. \begin{align} &\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b)^2}\\ &=\frac{a^2}{(x-y)^2}+\frac{(a+x)^2}{y^2}+\frac{(a+y)^2}{x^2}\\ &\ge\frac{x^2}{y^2}+\frac{y^2}{x^2}\\ & > 2 \end{align}