jmc

Since ${\omega }^3=1,$ $\frac{2}{\omega }=2{\omega }^2.$ Then multiplying both sides by $(a+\omega )(b+\omega )(c+\omega )(d+\omega ),$ we get $$(b+\omega )(c+\omega )(d+\omega )+(a+\omega )(c+\omega )(d+\omega )+(a+\omega )(b+\omega )(d+\omega )+(a+\omega )(b+\omega )(c+\omega )=2{\omega }^2(a+\omega )(b+\omega )(c+\omega )(d+\omega ).$$Expanding both sides, we get $$\begin{array}{rl}& 4{\omega }^3+3(a+b+c+d){\omega }^2+2(ab+ac+ad+bc+bd+cd)\omega +(abc+abd+acd+bcd)\\ & =2{\omega }^6+2(a+b+c+d){\omega }^5+2(ab+ac+ad+bc+bd+cd){\omega }^4+2(abc+abd+acd+bcd){\omega }^3+2abcd{\omega }^2.\end{array}$$Since ${\omega }^3=1,$ this simplifies to $$\begin{array}{rl}& 3(a+b+c+d){\omega }^2+2(ab+ac+ad+bc+bd+cd)\omega +(abc+abd+acd+bcd)+4\\ & =(2(a+b+c+d)+2abcd){\omega }^2+2(ab+ac+ad+bc+bd+cd)\omega +2(abc+abd+acd+bcd)+2.\end{array}$$Then $$(a+b+c+d-2abcd){\omega }^2-abc-abd-acd-bcd+2=0.$$Since ${\omega }^2$ is nonreal, we must have $a+b+c+d=2abcd.$ Then $abc+abd+acd+bcd=2.$

Hence, $$\begin{array}{rl}& \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}\\ & =\frac{(b+1)(c+1)(d+1)+(a+1)(c+1)(d+1)+(a+1)(b+1)(d+1)+(a+1)(b+1)(c+1)}{(a+1)(b+1)(c+1)(d+1)}\\ & =\frac{(abc+abd+acd+bcd)+2(ab+ac+ad+bc+bd+cd)+3(a+b+c+d)+4}{abcd+(abc+abd+acd+bcd)+(ab+ac+ad+bc+bd+cd)+(a+b+c+d)+1}\\ & =\frac{2+2(ab+ac+ad+bc+bd+cd)+6abcd+4}{abcd+2+(ab+ac+ad+bc+bd+cd)+2abcd+1}\\ & =\frac{6abcd+2(ab+ac+ad+bc+bd+cd)+6}{3abcd+(ab+ac+ad+bc+bd+cd)+3}\\ & =\boxed{2}.\end{array}$$