$ \dfrac{a}{b+c} + \dfrac{b}{a+c} + \dfrac{c}{a+b} \ge \dfrac{3}{2}$
$\to ( \dfrac{a}{b+c}+1) + (\dfrac{b}{a+c}+1) + ( \dfrac{c}{a+b}+1|) \ge \dfrac{3}{2} +3$
$\to \dfrac{a+b+c}{b+c} + \dfrac{a+b+c}{a+c} + \dfrac{a+b+c}{a+b} \ge \dfrac{9}{2}$
$\to (a+b+c)( \dfrac{1}{b+c} + \dfrac{1}{a+c} + \dfrac{1}{a+b}) \ge \dfrac{9}{2}$
Áp dụng Cauchy - Schwarz dạng Engel có
$ (a+b+c)( \dfrac{1}{b+c} + \dfrac{1}{a+c} + \dfrac{1}{a+b}) \ge (a+b+c)( \dfrac{(1+1+1)^2}{2(a+b+c)}) =$
$ = \dfrac{9(a+b+c)}{2(a+b+c)} = \dfrac{9}{2}$ (đpcm)
Dấu $=$ xảy ra khi $ a= b=c$
