If the equations $a(y+z)=x,b(z+x)=y,c(x+y)=z$  have non trivial solutions, then $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=$

 

$\left|\begin{array}{r}-1\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a\\ b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} -1 \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\\ c\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}c\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} -1\end{array}\right|=0$

Applying $C_2\to C_2-C_1,C_3\to C_3-C_1\text{ gives }$

$\left|\begin{array}{ccc}-1& a+1& a+1\\ b& -(b+1)& 0\\ c& 0& -(1+c)\end{array}\right|=0$

Applying $R_1\to \frac{R_1}{a+1},R_2\to \frac{R_2}{b+1},R_3\to \frac{R_3}{c+1}\text{ gives }$

$\left|\begin{array}{ccc}-\frac{1}{a+1}& 1& 1\\ \frac{b}{b+1}& -1& 0\\ \frac{c}{c+1}& 0& -1\end{array}\right|=0$

Expanding along $C_1$,we get

$\begin{array}{l}-\frac{1}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=0\\ \Rightarrow -\frac{1}{a+1}+1-\frac{1}{b+1}+1-\frac{1}{c+1}=0\\ \therefore \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2.\end{array}$