We have:\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{4}{a+b+c}\Leftrightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=4\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)\(\Rightarrow\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c\Leftrightarrow\dfrac{a^2+ab+ca}{b+c}+\dfrac{ba+b^2+cb}{c+a}+\dfrac{ac+cb+c^2}{a+b}=a+b+c\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{c^2}{a+b}+\dfrac{b^2}{a+c}+a+b+c=a+b+c\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0\)