Question

If \( a^2+b^2+c^2=1, a,b,c \in R \), then the set of extreme values of \(ab+bc+ca\) is

• A. \( \{-\frac{1}{2}, 2\} \)
• B. \( \{-1, 2\} \)
• C. \( \{-1, \frac{1}{2}\} \)
• D. \( \{-\frac{1}{2}, 1\} \)

Hint:

Use the identity \((a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)\). Since \((a+b+c)^2 \ge 0\), this gives \(ab+bc+ca \ge -(a^2+b^2+c^2)/2\).
Use the identity \(a^2+b^2+c^2 - (ab+bc+ca) = \frac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2) \ge 0\), which implies \(ab+bc+ca \le a^2+b^2+c^2\).

Answer 1

The Correct Option is D

We are given \(a^2+b^2+c^2=1\). We need to find the extreme values of \(S = ab+bc+ca\). Consider the identity: \((a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)\). Let \(X = a+b+c\). Then \(X^2 = 1 + 2S\). So, \(S = \frac{X^2-1}{2}\). Since \(X^2 = (a+b+c)^2 \ge 0\), we have \(2S+1 \ge 0 \Rightarrow S \ge -1/2\). This gives the minimum value of S as -1/2. This occurs when \(X^2=0 \Rightarrow a+b+c=0\). For example, if \(a=1/\sqrt{2}, b=-1/\sqrt{2}, c=0\), then \(a^2+b^2+c^2 = 1/2+1/2+0 = 1\). And \(ab+bc+ca = (1/\sqrt{2})(-1/\sqrt{2}) + 0 + 0 = -1/2\). For the maximum value: We know that \(a^2+b^2+c^2 - (ab+bc+ca) = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]\). Since squares are non-negative, \((a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0\). So, \(a^2+b^2+c^2 - (ab+bc+ca) \ge 0\). \(1 - S \ge 0 \Rightarrow S \le 1\). This gives the maximum value of S as 1. This occurs when \(a=b=c\). If \(a=b=c\), then \(a^2+a^2+a^2=1 \Rightarrow 3a^2=1 \Rightarrow a^2=1/3 \Rightarrow a = \pm 1/\sqrt{3}\). If \(a=b=c=1/\sqrt{3}\), then \(ab+bc+ca = 3a^2 = 3(1/3) = 1\). If \(a=b=c=-1/\sqrt{3}\), then \(ab+bc+ca = 3a^2 = 3(1/3) = 1\). So, the minimum value is -1/2 and the maximum value is 1. The set of extreme values is \(\{ -1/2, 1 \}\). This matches option (d). \[ \boxed{\{-\frac{1}{2}, 1\}} \]