Minimize the Expression 4

Notice that $\sum_{cyc}{\frac{bc}{(a-b)(a-c)}} = \frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)} = 1$

Thus, $\frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{a^2}{(a-b)^2} =$ $(\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b})^2 + 2\sum_{cyc}{\frac{bc}{(a-b)(a-c)}} = (\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b})^2 + 2 \geq 2$

Equality holds when $\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b} = 0$ .

Maybe this solution is wrong.

By Cauchy-Schwarz Inequality

$\left( \frac { a^{ 2 } }{ (b-c)^{ 2 } } +\frac { b^{ 2 } }{ (c-a)^{ 2 } } +\frac { c^{ 2 } }{ (a-b)^{ 2 } } \right) \left( { (b-c) }^{ 2 }+{ (c-a) }^{ 2 }+{ (a-b) }^{ 2 } \right) \ge { (a+b+c) }^{ 2 }\\ \left( \frac { a^{ 2 } }{ (b-c)^{ 2 } } +\frac { b^{ 2 } }{ (c-a)^{ 2 } } +\frac { c^{ 2 } }{ (a-b)^{ 2 } } \right) \ge \frac { { (a+b+c) }^{ 2 } }{ 2({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca) }$

Now we will prove another identity ${ (a+b+c) }^{ 2 }\ge ({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca)$ for $a,b,c \ge 0$ .

Consider

${ (a+b+c) }^{ 2 }-({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca)\\ ={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2(ab+bc+ca)-{ a }^{ 2 }-{ b }^{ 2 }-{ c }^{ 2 }+ab+bc+ca\\ =3(ab+bc+ca)\ge 0\quad (as\quad a,b,c\ge 0)$

Now the minimum value of $\frac { { (a+b+c) }^{ 2 } }{ 2({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca) }$ can be taken out as follows

${ (a+b+c) }^{ 2 }\ge ({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca)\\ \frac { { (a+b+c) }^{ 2 } }{ 2({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca) } \ge \frac { 1 }{ 2 }$

This means

$\left( \frac { a^{ 2 } }{ (b-c)^{ 2 } } +\frac { b^{ 2 } }{ (c-a)^{ 2 } } +\frac { c^{ 2 } }{ (a-b)^{ 2 } } \right) \left( { (b-c) }^{ 2 }+{ (c-a) }^{ 2 }+{ (a-b) }^{ 2 } \right) \ge { (a+b+c) }^{ 2 }\\ \left( \frac { a^{ 2 } }{ (b-c)^{ 2 } } +\frac { b^{ 2 } }{ (c-a)^{ 2 } } +\frac { c^{ 2 } }{ (a-b)^{ 2 } } \right) \ge \frac { { (a+b+c) }^{ 2 } }{ 2({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca) } \ge \frac { 1 }{ 2 }$

So, I thought let's round up bigger integers and entered $1,2$ and got the correct answer.