\(\frac{a}{ab+a+2016}+\frac{b}{bc+b+1}+\frac{2016c}{ac+2016c+2016}\)
\(=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{abc^2}{ac+abc^2+abc}\)
\(=\frac{a}{a.\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{abc^2}{ac.\left(1+bc+b\right)}\)
\(=\frac{1}{b+bc+1}+\frac{b}{b+bc+1}+\frac{bc}{b+bc+1}\)
\(=\frac{1+b+bc}{b+bc+1}=1\)