Inequality

[man111]

if $a,b,c$ are distinct real number, then minimum of $\displaystyle \frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b)^2}$

Using Cauchy Schwarz Inequality $\displaystyle \frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b)^2}\geq \frac{(a+b+c)^2}{(a-b)^2+(b-c)^2+(c-a)^2}$

can not go further, Help Required

[Sergic Primazon]

$$\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b)^2}-2=\dfrac{(a^3+b^3+c^3-(ab+bc+ca)(a+b+c)+6abc)^2}{(a-b)^2(b-c)^2(c-a)^2}$$