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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!

Given a = b + c, b = c + a, c = a + b, prove that 1/(a+1) + 1/(b+1) + 1/(c+1) = 1.

Explore the proof that 1/(a+1) + 1/(b+1) + 1/(c+1) equals 1 under the conditions a = b + c, b = c + a, and c = a + b. This algebraic problem is ideal for students in Grades 10-12 and demonstrates the application of symmetric equations and rational expressions.

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