\(abc+ab+bc+ca=4\)
\(\Leftrightarrow abc+2ab+2bc+2ca+4a+4b+4c+8=ab+bc+ca+4a+4b+4c+12\)
\(\Leftrightarrow\left(a+2\right)\left(b+2\right)\left(c+2\right)=\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)\)
\(\Leftrightarrow\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\)


