Let a, b, c be the lengths of the sides of a triangle (no two of them are equal) and `k in R`.If the roots of the equation `x^2 + 2(a+b+c)x+6k(ab+bc+ac)=0` are real, then:

Options

A. `l lt 2/3`
B. `k gt 2/3`
C. `k gt 1`
D. `k lt 1/4`

Answer

To solve the problem, we need to analyze the quadratic equation given and determine the conditions under which its roots are real. The equation is:

\[ x^2 + 2(a+b+c)x + 6k(ab + bc + ac) = 0 \]

### Step 1: Identify the coefficients
In the standard form of a quadratic equation \( ax^2 + bx + c = 0 \), we can identify:
- \( a = 1 \)
- \( b = 2(a + b + c) \)
- \( c = 6k(ab + bc + ac) \)

### Step 2: Apply the condition for real roots
For the roots of the quadratic equation to be real, the discriminant \( D \) must be non-negative:

\[ D = b^2 - 4ac \geq 0 \]

### Step 3: Calculate the discriminant
Substituting the values of \( a \), \( b \), and \( c \) into the discriminant:

\[ D = [2(a + b + c)]^2 - 4 \cdot 1 \cdot 6k(ab + bc + ac) \]

This simplifies to:

\[ D = 4(a + b + c)^2 - 24k(ab + bc + ac) \geq 0 \]

### Step 4: Rearranging the inequality
We can rearrange the inequality:

\[ 4(a + b + c)^2 \geq 24k(ab + bc + ac) \]

Dividing both sides by 4 gives:

\[ (a + b + c)^2 \geq 6k(ab + bc + ac) \]

### Step 5: Isolate \( k \)
Now, we can isolate \( k \):

\[ k \leq \frac{(a + b + c)^2}{6(ab + bc + ac)} \]

### Step 6: Analyze the triangle inequality
Since \( a, b, c \) are the sides of a triangle, we can use the triangle inequality which states that for any triangle:

1. \( a + b > c \)
2. \( a + c > b \)
3. \( b + c > a \)

From these inequalities, we can derive that:

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

### Step 7: Substitute and simplify
Using the triangle inequality, we know that \( a^2 + b^2 + c^2 < 2(ab + ac + bc) \). Thus, we can say:

\[ (a + b + c)^2 < 4(ab + ac + bc) \]

### Step 8: Substitute back into the inequality for \( k \)
Now substituting this back into our inequality for \( k \):

\[ k \leq \frac{4(ab + ac + bc)}{6(ab + ac + bc)} = \frac{2}{3} \]

### Conclusion
Thus, we conclude that:

\[ k < \frac{2}{3} \]