Đặt:
$\begin{cases}b + c - a = x\\a + c - b = y\\a + b - c =z\end{cases}$
$\Leftrightarrow \begin{cases}a = \dfrac{y+z}{2}\\b = \dfrac{x + z}{2}\\c = \dfrac{x + y}{2}\end{cases}$
Ta được:
$\dfrac{a}{b + c -a} + \dfrac{b}{a + c - b} + \dfrac{c}{a + b - c}$
$= \dfrac{y + z}{2x} + \dfrac{x + z}{2y} + \dfrac{x + y}{2z}$
$= \dfrac{1}{2}\left(\dfrac{y}{x} + \dfrac{z}{x} + \dfrac{x}{y} + \dfrac{z}{y} + \dfrac{x}{z} + \dfrac{y}{z}\right)$
$\geq \dfrac{1}{2}\left(2\sqrt{\dfrac{y}{x}\cdot\dfrac{x}{y}} + 2\sqrt{\dfrac{z}{x}\cdot\dfrac{x}{z}} + 2\sqrt{\dfrac{z}{y}\cdot\dfrac{y}{z}}\right) = \dfrac{1}{2}.6 = 3$
Dấu = xảy ra $\Leftrightarrow x = y = z \Leftrightarrow a = b = c$
Vậy $\dfrac{a}{b + c -a} + \dfrac{b}{a + c - b} + \dfrac{c}{a + b - c} \geq 3$ tại $a = b = c$
