Áp dụng AM-GM:
\(\dfrac{1}{a\left(a+b\right)}+\dfrac{1}{b\left(b+c\right)}+\dfrac{1}{c\left(c+a\right)}\ge\dfrac{3}{\sqrt[3]{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\dfrac{3}{\sqrt[3]{\left(ab+bc\right)\left(bc+ac\right)\left(ac+ab\right)}}\ge\dfrac{3}{\dfrac{1}{3}.2\left(ab+bc+ca\right)}\ge\dfrac{27}{2\left(a+b+c\right)^2}\)