If a, b, c are sides of a triangle, then `((a+b+c)^(2))/(ab+bc+ca)` always belongs to

Options

A. [1,2]
B. [2,3]
C. [3,4]
D. [4,5]

Explanation

To solve the problem, we need to find the range of the expression \(\frac{(a+b+c)^2}{ab+bc+ca}\) where \(a\), \(b\), and \(c\) are the sides of a triangle.

### Step-by-Step Solution:

1. **Apply the Cauchy-Schwarz Inequality**:
We can use the Cauchy-Schwarz inequality, which states that for any non-negative real numbers \(x_1, x_2, \ldots, x_n\) and \(y_1, y_2, \ldots, y_n\):
\[ (x_1^2 + x_2^2 + \ldots + x_n^2)(y_1^2 + y_2^2 + \ldots + y_n^2) \geq (x_1y_1 + x_2y_2 + \ldots + x_ny_n)^2 \]
In our case, let \(x_1 = a\), \(x_2 = b\), \(x_3 = c\) and \(y_1 = b+c\), \(y_2 = c+a\), \(y_3 = a+b\). Then we have:
\[ (a^2 + b^2 + c^2)((b+c)^2 + (c+a)^2 + (a+b)^2) \geq (ab + ac + bc)^2 \]

2. **Simplifying the Expression**:
From the triangle inequality, we know:
\[ a^2 + b^2 + c^2 \geq ab + ac + bc \]
Therefore, we can conclude:
\[ (a+b+c)^2 \geq 3(ab + ac + bc) \]
This implies:
\[ \frac{(a+b+c)^2}{ab + ac + bc} \geq 3 \]
This gives us the lower bound of our expression.

3. **Finding the Upper Bound**:
To find the upper bound, we can use the fact that in a triangle, the sides satisfy the triangle inequality, which implies:
\[ a + b > c, \quad b + c > a, \quad c + a > b \]
By squaring these inequalities and adding them, we can derive:
\[ (a+b+c)^2 < 4(ab + ac + bc) \]
Thus, we can conclude:
\[ \frac{(a+b+c)^2}{ab + ac + bc} < 4 \]

4. **Conclusion**:
Combining both results, we have:
\[ 3 \leq \frac{(a+b+c)^2}{ab + ac + bc} < 4 \]

Thus, the final result is that the expression \(\frac{(a+b+c)^2}{ab + ac + bc}\) always belongs to the interval \([3, 4)\).