Question : If $a^{2}=b+c,b^{2}=c+a,c^{2}=a+b$, the value of $3\left (\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)$ is:
Option 1: $1$
Option 2: $\frac{1}{3}$
Option 3: $3$
Option 4: $4$
Correct Answer: $3$
Solution :
Given: $a^{2}=b+c,b^{2}=c+a$, and $c^{2}=a+b$
Now, $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$
= $\frac{a}{a^2+a}+\frac{b}{b^2+b}+\frac{c}{c^2+c}$ (multiplying the numerator and the denominator of each term by $a,b$, and $c$ respectively)
= $\frac{a}{b+c+a}+\frac{b}{c+a+b}+\frac{c}{a+b+c}$ (putting the values of $a^2,b^2$ and $c^2$)
= $\frac{a+b+c}{a+b+c}$
= $1$
So, $3\left (\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)=3×1=3$
Hence, the correct answer is $3$.
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