If a,b,c are the sides of triangle then `(a)/(c +a -b) + (b)/(a +b -c) + (c )/(b +c-a)` can take vlaue (s)

Options

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

Answer

To solve the problem, we need to evaluate the expression:

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

where \(a\), \(b\), and \(c\) are the sides of a triangle.

### Step 1: Understand the Conditions
Since \(a\), \(b\), and \(c\) are the sides of a triangle, they must satisfy the triangle inequalities:
1. \(a + b > c\)
2. \(b + c > a\)
3. \(c + a > b\)

From these inequalities, we can conclude that:
- \(c + a - b > 0\)
- \(a + b - c > 0\)
- \(b + c - a > 0\)

This means that all the denominators in our expression are positive.

### Step 2: Apply the AM-GM Inequality
We can apply the Arithmetic Mean - Geometric Mean (AM-GM) inequality to the expression. According to AM-GM, for any positive numbers \(x_1, x_2, x_3\):
\[ \frac{x_1 + x_2 + x_3}{3} \geq \sqrt[3]{x_1 x_2 x_3} \]

Let:
- \(x_1 = \frac{a}{c + a - b}\)
- \(x_2 = \frac{b}{a + b - c}\)
- \(x_3 = \frac{c}{b + c - a}\)

Then, we can write:
\[ \frac{x_1 + x_2 + x_3}{3} \geq \sqrt[3]{x_1 x_2 x_3} \]

### Step 3: Find the Common Denominator
To combine the fractions, we need a common denominator:
\[ \text{Common Denominator} = (c + a - b)(a + b - c)(b + c - a) \]

### Step 4: Evaluate the Expression
Now we can write:
\[ \frac{a(a + b - c)(b + c - a) + b(b + c - a)(c + a - b) + c(c + a - b)(a + b - c)}{(c + a - b)(a + b - c)(b + c - a)} \]

### Step 5: Analyze the Numerator
We need to analyze the numerator:
\[ N = a(a + b - c)(b + c - a) + b(b + c - a)(c + a - b) + c(c + a - b)(a + b - c) \]
This expression can be simplified, but it is complex. However, we know that since \(a\), \(b\), and \(c\) are positive, \(N\) will also be positive.

### Step 6: Conclude the Value
Since both the numerator and denominator are positive, the entire expression is positive.

### Final Result
The expression can take values greater than or equal to 1, and it can be shown that the minimum value is 1 when \(a = b = c\).