If a, b and c are distinct positive real numbers and `a^2 + b^2 + c^2 = 1,` then ab + bc + ca is

Options

A. less than 1
B. equal to 1
C. greater than 1
D. any real number

Answer

To solve the problem, we need to find the value of \( ab + bc + ca \) given that \( a^2 + b^2 + c^2 = 1 \) and \( a, b, c \) are distinct positive real numbers.

### Step-by-step Solution:

1. **Start with the identity**:
We know from algebra that:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]

2. **Substitute the known value**:
Given \( a^2 + b^2 + c^2 = 1 \), we can substitute this into the identity:
\[ (a + b + c)^2 = 1 + 2(ab + ac + bc) \]

3. **Rearranging the equation**:
Rearranging gives us:
\[ 2(ab + ac + bc) = (a + b + c)^2 - 1 \]
Therefore, we can express \( ab + ac + bc \) as:
\[ ab + ac + bc = \frac{(a + b + c)^2 - 1}{2} \]

4. **Analyzing the expression**:
Since \( a, b, c \) are distinct positive real numbers, \( a + b + c \) must be greater than 1 (because if all were less than or equal to 1, their squares would sum to less than or equal to 1, contradicting the distinctness and positivity).

5. **Conclusion**:
Thus, \( (a + b + c)^2 - 1 > 0 \) implies:
\[ ab + ac + bc > 0 \]
However, we need to determine if \( ab + ac + bc < 1 \), \( = 1 \), or \( > 1 \).

Since \( a^2 + b^2 + c^2 = 1 \) and \( a, b, c \) are distinct positive numbers, we can infer that:
\[ ab + ac + bc < \frac{1}{2} \quad \text{(by Cauchy-Schwarz inequality)} \]
Therefore, \( ab + ac + bc < 1 \).

### Final Answer:
Thus, \( ab + ac + bc < 1 \).

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