Đặt: $P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b};Q=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}$
Ta có:
$P=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3$
$=(a+b+c)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3$
$\ge(a+b+c).\dfrac{9}{2(a+b+c)}-3=\dfrac{3}{2}$
$Q=\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a}$
$\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}+2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=6$
Suy ra: $P+Q\ge\dfrac{15}{2}$
Dấu bằng xảy ra khi: $a=b=c$