If $\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1$, then find the value of $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$.

Take the first component expression:
$$\frac{a^2}{b+c}=1⟹a^2=b+c$$

Add $a$ to both sides of the equation:

$$a^2+a=a+b+c$$

$$a(1+a)=a+b+c⟹\frac{1}{1+a}=\frac{a}{a+b+c}$$

Using the exact same algebraic step for the other symmetric terms:

$$\frac{1}{1+b}=\frac{b}{a+b+c}$$

$$\frac{1}{1+c}=\frac{c}{a+b+c}$$

Summing all three rewritten fractions gives:

$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1$$