If a,b,c are the sides of a triangle then `a/(b+c-1)+b/(c+a-b)+c/(a+b-c)=`

Options

A. `gt3`
B. `lt3`
C. `le2`
D. `ge2`

Explanation

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

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

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

### Step 1: Set up the expression

Let:
\[ S = \frac{a}{b+c-1} + \frac{b}{c+a-1} + \frac{c}{a+b-1} \]

### Step 2: Multiply the entire expression by 2

To simplify our calculations, we can multiply the entire expression by 2:
\[ 2S = 2 \left( \frac{a}{b+c-1} + \frac{b}{c+a-1} + \frac{c}{a+b-1} \right) \]

### Step 3: Add 3 to the right-hand side

Next, we add 3 to the right-hand side:
\[ 2S + 3 = 2 \left( \frac{a}{b+c-1} + \frac{b}{c+a-1} + \frac{c}{a+b-1} \right) + 3 \]

### Step 4: Rewrite the expression

We can rewrite the expression:
\[ 2S + 3 = 2 \left( \frac{a}{b+c-1} + 1 \right) + 2 \left( \frac{b}{c+a-1} + 1 \right) + 2 \left( \frac{c}{a+b-1} + 1 \right) - 3 \]

### Step 5: Combine terms

Combining the terms, we have:
\[ 2S + 3 = \frac{2a + (b+c-1)}{b+c-1} + \frac{2b + (c+a-1)}{c+a-1} + \frac{2c + (a+b-1)}{a+b-1} \]

### Step 6: Apply the AM-GM inequality

Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we know that:
\[ \frac{x_1 + x_2 + x_3}{3} \geq \sqrt[3]{x_1 x_2 x_3} \]
where \(x_1 = \frac{1}{b+c-a}\), \(x_2 = \frac{1}{c+a-b}\), \(x_3 = \frac{1}{a+b-c}\).

### Step 7: Establish the inequality

From the AM-GM inequality, we can establish that:
\[ \frac{1}{b+c-a} + \frac{1}{c+a-b} + \frac{1}{a+b-c} \geq \frac{9}{(b+c-a) + (c+a-b) + (a+b-c)} \]
which simplifies to:
\[ \frac{9}{a+b+c} \]

### Step 8: Solve for \(S\)

Thus, we can conclude:
\[ 2S \geq 9 \]
This implies:
\[ S \geq \frac{9}{2} \]

### Step 9: Conclusion

Therefore, the final result is:
\[ \frac{a}{b+c-1} + \frac{b}{c+a-1} + \frac{c}{a+b-1} \geq \frac{9}{2} \]