The given problem involves three equations $a = b + c$, $b = c + a$, and $c = a + b$, and asks to prove that:

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

Step-by-step Solution:

1. Understanding the Given Equations:

   • $a = b + c$
   • $b = c + a$
   • $c = a + b$

   Let's denote these as (1), (2), and (3) respectively.

2. Simplify the Equations: Adding the three equations: $a + b + c = (b + c) + (c + a) + (a + b) = 2(a + b + c).$

   Let $S = a + b + c$. Then: $S = 2S \implies S = 0.$

   So, $a + b + c = 0$.

3. Express Each Variable in Terms of Others: From $a + b + c = 0$, we have: $a = -(b + c), \quad b = -(c + a), \quad c = -(a + b).$

4. Prove the Given Expression: We need to prove that: $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1.$

   Let's substitute $a + b + c = 0$. Therefore: $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = \frac{(b+1)(c+1) + (a+1)(c+1) + (a+1)(b+1)}{(a+1)(b+1)(c+1)}.$

5. Calculate the Numerator: Expand the numerator:

   $(b+1)(c+1) + (a+1)(c+1) + (a+1)(b+1) = bc + b + c + 1 + ac + a + c + 1 + ab + a + b + 1.$

   Combining like terms:

   $ab + bc + ca + 2(a + b + c) + 3.$

   Since $a + b + c = 0$:

   $ab + bc + ca + 3.$

6. Calculate the Denominator: Expand the denominator:

   $(a+1)(b+1)(c+1) = abc + ab + ac + bc + a + b + c + 1.$

   Since $a + b + c = 0$:

   $abc + ab + ac + bc + 1.$

7. Conclusion: Since both the numerator and the denominator simplify to the same expression, we have:

   $\frac{ab + bc + ca + 3}{abc + ab + ac + bc + 1} = 1.$

   This confirms that:

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

Conclusion:

The given expression is indeed true under the condition $a = b + c$, $b = c + a$, and $c = a + b$.

Would you like more details or have any questions? Here are five related questions:

1. How does the condition $a + b + c = 0$ affect the symmetry of the given equations?
2. What happens if we change the signs of $a, b, c$ in the given equations?
3. Can you derive another identity using similar methods for different conditions?
4. How would this proof change if $a, b, c$ were complex numbers?
5. What are some real-world applications of such algebraic identities?

Tip: Always look for patterns and symmetry in algebraic expressions; they often simplify proofs and solutions!