If `(1)/(a+1) + (1)/(1+b) + (1)/(1 + c) =1`
find `(a)/(1+a) + (b)/(1+b) + (c )/(1+c)`= ?

Answer

To solve the equation

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

and find

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

we can follow these steps:

### Step 1: Rewrite the given equation
We start with the equation:

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

### Step 2: Find a common denominator
The common denominator for the left-hand side is \((a+1)(1+b)(1+c)\). Therefore, we can rewrite the equation as:

\[ \frac{(1+b)(1+c) + (a+1)(1+c) + (a+1)(1+b)}{(a+1)(1+b)(1+c)} = 1. \]

### Step 3: Simplify the numerator
Now, we will simplify the numerator:

\[ (1+b)(1+c) + (a+1)(1+c) + (a+1)(1+b). \]

Expanding each term:

1. \((1+b)(1+c) = 1 + b + c + bc\)
2. \((a+1)(1+c) = a + ac + 1 + c\)
3. \((a+1)(1+b) = a + ab + 1 + b\)

Combining these, we have:

\[ (1 + b + c + bc) + (a + ac + 1 + c) + (a + ab + 1 + b). \]

This simplifies to:

\[ 2 + 2a + 2b + 2c + ab + ac + bc. \]

### Step 4: Set the numerator equal to the denominator
Setting the numerator equal to the denominator gives:

\[ 2 + 2a + 2b + 2c + ab + ac + bc = (a+1)(1+b)(1+c). \]

### Step 5: Find the expression we need
Now, we need to find:

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

We can rewrite each term:

\[ \frac{a}{1+a} = 1 - \frac{1}{1+a}, \quad \frac{b}{1+b} = 1 - \frac{1}{1+b}, \quad \frac{c}{1+c} = 1 - \frac{1}{1+c}. \]

Thus,

\[ \frac{a}{1+a} + \frac{b}{1+b} + \frac{c}{1+c} = 3 - \left(\frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c}\right). \]

### Step 6: Substitute the known value
From the original equation, we know:

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

Therefore,

\[ \frac{a}{1+a} + \frac{b}{1+b} + \frac{c}{1+c} = 3 - 1 = 2. \]

### Final Answer
Thus, the value of

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

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