$A = \dfrac{a}{b + c - a} + \dfrac{b}{c + a - b} + \dfrac{c}{a + b - c}$
$\to A = \dfrac{a^2}{ab + ac - a^2} + \dfrac{b^2}{bc + ab- b^2} + \dfrac{c^2}{ac + bc - c^2}$
With Engel's form from Cauchy-Schwarz inequality, we have:
$A \geq \dfrac{(a + b + c)^2}{2(ab + bc + ca) - (a^2 + b^2 + c^2)}$
We also have:
$ab + bc + ca \leq \dfrac{(a + b + c)^2}{3}$
$\Leftrightarrow 2(ab + bc + ca)\leq \dfrac{2}{3}(a + b + c)^2$
and $(a + b + c)^2 \leq 3(a^2 + b^2 + c^2)$
$\Leftrightarrow -(a^2 + b^2 + c^2) \leq -\dfrac{1}{3}(a + b+ c)^2$
Therefore:
$2(ab + bc + ca) - (a^2 + b^2 + c^2) \leq \dfrac{2}{3}(a + b + c)^2 -\dfrac{1}{3}(a + b + c)^2=\dfrac{1}{3}(a + b + c)^2$
$\Leftrightarrow \dfrac{(a + b + c)^2}{2(ab + bc + ca) - (a^2 + b^2 + c^2)} \geq \dfrac{(a + b + c)^2}{\dfrac{1}{3}(a + b + c)^2}=3$
With equality if and only if: $a = b = c$
