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

Options

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

Explanation

To find the interval in which \( ab + bc + ca \) lies given that \( a^2 + b^2 + c^2 = 1 \), we can follow these steps:

### Step 1: Write down the given condition
We start with the equation:
\[ a^2 + b^2 + c^2 = 1 \]

### Step 2: Use the identity for the square of a sum
We know that:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]
Substituting the given condition into this identity, we get:
\[ (a + b + c)^2 = 1 + 2(ab + bc + ca) \]

### Step 3: Analyze the expression
Since \( (a + b + c)^2 \geq 0 \) (the square of a real number is non-negative), we can write:
\[ 1 + 2(ab + bc + ca) \geq 0 \]

### Step 4: Rearranging the inequality
Rearranging the above inequality gives us:
\[ 2(ab + bc + ca) \geq -1 \]
Dividing both sides by 2, we find:
\[ ab + bc + ca \geq -\frac{1}{2} \]

### Step 5: Finding an upper bound
Next, we can use the Cauchy-Schwarz inequality:
\[ (a^2 + b^2 + c^2)(1^2 + 1^2 + 1^2) \geq (a + b + c)^2 \]
Substituting \( a^2 + b^2 + c^2 = 1 \) into the inequality gives:
\[ 1 \cdot 3 \geq (a + b + c)^2 \]
Thus,
\[ (a + b + c)^2 \leq 3 \]
Taking the square root, we have:
\[ |a + b + c| \leq \sqrt{3} \]
Now, substituting back into our earlier equation:
\[ 1 + 2(ab + bc + ca) \leq 3 \]
This simplifies to:
\[ 2(ab + bc + ca) \leq 2 \]
Dividing by 2 gives:
\[ ab + bc + ca \leq 1 \]

### Step 6: Conclusion
Combining both inequalities, we conclude that:
\[ -\frac{1}{2} \leq ab + bc + ca \leq 1 \]
Thus, the interval in which \( ab + bc + ca \) lies is:
\[ \left[-\frac{1}{2}, 1\right] \]