Question

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

• A. \(\left\{\dfrac{1}{2},2\right\}\)
• B. \(\{-1,2\}\)
• C. \(\left\{-1,\dfrac{1}{2}\right\}\)
• D. \(\left\{-\dfrac{1}{2},1\right\}\)

Hint:

For expressions like \(ab+bc+ca\), use the identity \[ (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca). \] It helps in finding maximum and minimum values easily.

Answer 1

The Correct Option is D

Step 1: Use the identity \((a+b+c)^2 = 1 + 2(ab+bc+ca)\).
Since \((a+b+c)^2 \geq 0\), we have \(ab+bc+ca \geq -\tfrac{1}{2}\). Equality holds when \(a+b+c=0\), e.g. \(a=1,b=0,c=-1\) (check: \(1+0+1=2\neq1\)); more precisely \(a=\tfrac{1}{\sqrt2}, b=0, c=-\tfrac{1}{\sqrt2}\) gives \(ab+bc+ca=-\tfrac{1}{2}\).

Step 2: Find the maximum.
By AM inequality, \((a+b+c)^2 \leq 3(a^2+b^2+c^2)=3\), so \(ab+bc+ca \leq 1\). Equality when \(a=b=c=\tfrac{1}{\sqrt3}\).

Step 3: State the extreme values.
The set of extreme values is \(\left\{-\tfrac{1}{2},\, 1\right\}\).
\[\boxed{\left\{-\frac{1}{2}, 1\right\}}\]