Đặt $b+c-a=x$, $a+c-b=y$, $a+b-c=z$
$\Rightarrow b+c-a+a+c-b=x+y\Rightarrow 2c=x+y\Rightarrow c=\dfrac{x+y} 2$
Tương tự $a=\dfrac{y+z} 2, b=\dfrac{x+z} 2$
Do đó:
$\begin{array}{l}
\dfrac{a}{{b + c - a}} + \dfrac{b}{{a + c - b}} + \dfrac{c}{{a + b - c}} = \dfrac{{y + z}}{{2x}} + \dfrac{{x + z}}{{2y}} + \dfrac{{x + y}}{{2z}}\\
= \dfrac{1}{2}\left( {\dfrac{x}{y} + \dfrac{y}{x} + \dfrac{y}{z} + \dfrac{z}{y} + \dfrac{z}{x} + \dfrac{x}{z}} \right) \ge \dfrac{1}{2}\left( {2 + 2 + 2} \right) = 3\\
\Rightarrow \dfrac{a}{{b + c - a}} + \dfrac{b}{{a + c - b}} + \dfrac{c}{{a + b - c}} \ge 3\\
' = ' \Leftrightarrow x = y = z \Leftrightarrow a = b = c
\end{array}$
