Step 1: Understanding the Concept:
We can find the range of \( ab + bc + ca \) using the expansion of \( (a+b+c)^2 \) and the fact that squares of real numbers are non-negative.
: Key Formula or Approach:
1. \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \ge 0 \).
2. \( \frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2] \ge 0 \).
Step 2: Detailed Explanation:
Lower Bound:
We know that for any real numbers \( a, b, c \):
\[ (a + b + c)^2 \ge 0 \]
\[ a^2 + b^2 + c^2 + 2(ab + bc + ca) \ge 0 \]
Given \( a^2 + b^2 + c^2 = 1 \):
\[ 1 + 2(ab + bc + ca) \ge 0 \implies ab + bc + ca \ge -1/2 \]
Upper Bound:
We know that:
\[ (a - b)^2 + (b - c)^2 + (c - a)^2 \ge 0 \]
\[ 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca \ge 0 \]
\[ 2(a^2 + b^2 + c^2) \ge 2(ab + bc + ca) \]
Substitute \( a^2 + b^2 + c^2 = 1 \):
\[ 2(1) \ge 2(ab + bc + ca) \implies ab + bc + ca \le 1 \]
Combining these results, the value lies in \( [-1/2, 1] \).
Step 3: Final Answer:
The interval is \( [-1/2, 1] \).