If `(a^(2))/( b + c ) = (b ^(2))/(c + a) = (c ^(2))/( a + b) = 1 ` then ` (1)/( 1 + a ) + (1)/( 1 + b ) + (1)/(a + c)` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we start with the given equations:

\[ \frac{a^2}{b+c} = \frac{b^2}{c+a} = \frac{c^2}{a+b} = 1 \]

This implies:

1. \( a^2 = b + c \)
2. \( b^2 = c + a \)
3. \( c^2 = a + b \)

From these equations, we can express \( b+c \), \( c+a \), and \( a+b \) in terms of \( a^2 \), \( b^2 \), and \( c^2 \) respectively.

### Step 1: Set the equations equal to 1

Since all three expressions equal 1, we can write:

\[ a^2 = b + c \]
\[ b^2 = c + a \]
\[ c^2 = a + b \]

### Step 2: Analyze the symmetry

Notice that the equations are symmetric in \( a \), \( b \), and \( c \). This suggests that \( a = b = c \). Let's denote \( a = b = c = k \).

### Step 3: Substitute \( a = b = c = k \)

Substituting into the first equation:

\[ k^2 = k + k \implies k^2 = 2k \]

### Step 4: Solve for \( k \)

Rearranging gives:

\[ k^2 - 2k = 0 \]

Factoring out \( k \):

\[ k(k - 2) = 0 \]

Thus, \( k = 0 \) or \( k = 2 \). Since we are looking for positive values, we take \( k = 2 \).

### Step 5: Calculate the desired expression

Now we need to compute:

\[ \frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c} \]

Substituting \( a = b = c = 2 \):

\[ \frac{1}{1 + 2} + \frac{1}{1 + 2} + \frac{1}{2 + 2} = \frac{1}{3} + \frac{1}{3} + \frac{1}{4} \]

### Step 6: Simplify the expression

Calculating this gives:

\[ \frac{1}{3} + \frac{1}{3} + \frac{1}{4} = \frac{2}{3} + \frac{1}{4} \]

To add these fractions, we find a common denominator, which is 12:

\[ \frac{2}{3} = \frac{8}{12}, \quad \frac{1}{4} = \frac{3}{12} \]

Thus,

\[ \frac{8}{12} + \frac{3}{12} = \frac{11}{12} \]

### Final Answer

The final value of the expression is:

\[ \frac{11}{12} \]