If `|{:(,b+c,c+a,a+b),(,c+a,a+b,b+c),(,a+b,b+c,c+a):}|ge 0, "where a ,b,c,"inR,"then"(a+b)/(c)"is"`

Answer

To solve the problem, we need to evaluate the determinant given by:

\[ D = \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \]

and determine the condition under which \( \frac{a+b}{c} \) is evaluated.

### Step 1: Apply Row Operation
We can simplify the determinant by adding all three rows together and replacing the first row with this sum.

\[ R_1 \rightarrow R_1 + R_2 + R_3 \]

This gives us:

\[ D = \begin{vmatrix} 2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Now, we can factor out \( 2(a+b+c) \) from the first row:

\[ D = 2(a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \]

### Step 3: Apply Column Operations
Next, we can simplify the determinant further by performing column operations. We will subtract the first column from the second and third columns:

\[ C_2 \rightarrow C_2 - C_1 \quad \text{and} \quad C_3 \rightarrow C_3 - C_1 \]

This results in:

\[ D = 2(a+b+c) \begin{vmatrix} 1 & 0 & 0 \\ c+a - (b+c) & (a+b) - (b+c) & (b+c) - (c+a) \\ (a+b) - (b+c) & (b+c) - (c+a) & (c+a) - (a+b) \end{vmatrix} \]

This simplifies to:

\[ D = 2(a+b+c) \begin{vmatrix} 1 & 0 & 0 \\ -a & a-b & b-c \\ b-a & c-b & a-c \end{vmatrix} \]

### Step 4: Evaluate the Determinant
The determinant now can be evaluated as:

\[ D = 2(a+b+c) \cdot 1 \cdot \begin{vmatrix} -a & a-b \\ b-a & c-b \end{vmatrix} \]

Calculating this 2x2 determinant:

\[ D = 2(a+b+c) \left[ -a(c-b) - (a-b)(b-a) \right] \]

### Step 5: Set the Determinant Greater Than or Equal to Zero
Now we need to ensure that \( D \geq 0 \). This leads us to the conclusion that:

\[ -a(c-b) - (a-b)(b-a) \geq 0 \]

### Step 6: Analyze the Condition
The condition implies that \( a, b, c \) must be equal for the determinant to be non-negative. Thus, we can set \( a = b = c \).

### Step 7: Calculate \( \frac{a+b}{c} \)
If \( a = b = c \), then:

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

### Final Answer
Thus, we conclude that:

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