$\begin{array}{l}\dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b}\\ = \dfrac{a}{b+c} +1 + \dfrac{b}{c+a} +1 + \dfrac{c}{a+b} + 1 -3\\ =\dfrac{a+b+c}{b+c} + \dfrac{a+b+c}{c+a} + \dfrac{a+b+c}{a+b} -3\\ =(a+b+c)\left(\dfrac{1}{b+c} +\dfrac{1}{c+a} +\dfrac{1}{a+b} \right) -3\\ \geq (a+b+c)\cdot\dfrac{(1+1+1)^2}{2(a+b+c)} - 3 =\dfrac{3}{2}\\ \text{Dấu = xảy ra}\,\,\Leftrightarrow a = b = c \end{array}$
