Question

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

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

Hint:

Minimum at \(a+b+c=0\), maximum at \(a=b=c\) for symmetric expressions.

Answer 1

The Correct Option is C

Concept: \[ (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca) \]

Step 1: Rewrite expression.
\[ ab+bc+ca = \frac{(a+b+c)^2 - 1}{2} \]

Step 2: Minimum value.
Using identity: \[ a^2 + b^2 + c^2 \ge ab + bc + ca \] \[ \Rightarrow 1 \ge ab+bc+ca \] Also, \[ (a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0 \Rightarrow a^2+b^2+c^2 \ge ab+bc+ca \] For minimum, take \(a+b+c = 0\): \[ ab+bc+ca = \frac{0 - 1}{2} = -\frac{1}{2} \]

Step 3: Maximum value.
Occurs when \(a=b=c\): \[ 3a^2 = 1 \Rightarrow a = \frac{1}{\sqrt{3}} \] \[ ab+bc+ca = 3a^2 = 1 \]

Step 4: Final interval.
\[ -\frac{1}{2} \le ab+bc+ca \le 1 \]