If`a^(2)+b^(2)+c^(2)=1` then `ab+bc+ca` lies in the interval

Options

A. `[1,2]`
B. `[0,1/2)`
C. `[-1/2,1]`
D. `[0,1]`

Answer

To solve the problem, we need to determine the range of the expression \( ab + bc + ca \) given the constraint \( a^2 + b^2 + c^2 = 1 \).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We start with the condition:
\[ a^2 + b^2 + c^2 = 1 \]

2. **Using the Cauchy-Schwarz Inequality**:
By applying the Cauchy-Schwarz inequality, we can derive that:
\[ (a + b + c)^2 \leq (1^2 + 1^2 + 1^2)(a^2 + b^2 + c^2) = 3(a^2 + b^2 + c^2) = 3 \]
This implies:
\[ a + b + c \leq \sqrt{3} \]

3. **Finding the Lower Bound**:
We can also use the identity:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]
Rearranging gives:
\[ ab + bc + ca = \frac{(a + b + c)^2 - (a^2 + b^2 + c^2)}{2} \]
Substituting \( a^2 + b^2 + c^2 = 1 \):
\[ ab + bc + ca = \frac{(a + b + c)^2 - 1}{2} \]

4. **Finding the Upper Bound**:
The maximum value of \( (a + b + c)^2 \) is 3 (from Cauchy-Schwarz), thus:
\[ ab + bc + ca \leq \frac{3 - 1}{2} = 1 \]

5. **Finding the Lower Bound**:
To find the minimum value, we can use the fact that:
\[ (a - b)^2 + (b - c)^2 + (c - a)^2 \geq 0 \]
Expanding this gives:
\[ 2(a^2 + b^2 + c^2) - 2(ab + bc + ca) \geq 0 \]
Substituting \( a^2 + b^2 + c^2 = 1 \):
\[ 2 - 2(ab + bc + ca) \geq 0 \implies ab + bc + ca \leq 1 \]
Rearranging gives:
\[ ab + bc + ca \geq -\frac{1}{2} \]

6. **Final Conclusion**:
Combining both bounds, we find:
\[ -\frac{1}{2} \leq ab + bc + ca \leq 1 \]
Therefore, the interval for \( ab + bc + ca \) is:
\[ \boxed{[-\frac{1}{2}, 1]} \]