If a, b and c represent the lengths of sides of a triangle then the possible integeral value of `(a)/(b+c) + (b)/(c+a) + (c)/(a +b)` is _____

Explanation

To find the possible integral value of the expression \( \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \) where \( a, b, c \) are the lengths of the sides of a triangle, we can follow these steps:

### Step 1: Understand the properties of triangle sides
In any triangle, the sum of the lengths of any two sides must be greater than the length of the third side. Therefore, we have the following inequalities:
1. \( a + b > c \)
2. \( a + c > b \)
3. \( b + c > a \)

### Step 2: Analyze the expression
We need to evaluate the expression:
\[ S = \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \]

### Step 3: Use inequalities
From the properties of triangles, we can derive that:
\[ \frac{a}{b+c} < 1, \quad \frac{b}{c+a} < 1, \quad \frac{c}{a+b} < 1 \]
Thus, we can conclude:
\[ S < 3 \]

### Step 4: Establish a lower bound
To establish a lower bound, we can use the Cauchy-Schwarz inequality:
\[ \left( \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \right) \left( (b+c) + (c+a) + (a+b) \right) \geq (a+b+c)^2 \]
The left-hand side simplifies to:
\[ (b+c) + (c+a) + (a+b) = 2(a+b+c) \]
Thus, we have:
\[ S \cdot 2(a+b+c) \geq (a+b+c)^2 \]
This leads to:
\[ S \geq \frac{(a+b+c)^2}{2(a+b+c)} = \frac{a+b+c}{2} \]

### Step 5: Determine the range of \( S \)
Since \( a, b, c \) are positive integers, \( a+b+c \) is at least 3 (the minimum sum of the sides of a triangle). Therefore:
\[ S \geq \frac{3}{2} = 1.5 \]

### Step 6: Combine the bounds
From the analysis, we have:
\[ 1.5 < S < 3 \]
The possible integral values of \( S \) that lie between \( 1.5 \) and \( 3 \) is only:
\[ 2 \]

### Final Answer
Thus, the possible integral value of \( \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \) is:
\[ \boxed{2} \]