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 involving \(a+b+c\).
We know that \[ (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca) \] Given, \[ a^2+b^2+c^2=1 \] Therefore, \[ (a+b+c)^2=1+2(ab+bc+ca) \]

Step 2: Find the maximum value.
By Cauchy-Schwarz inequality, \[ (a+b+c)^2\leq 3(a^2+b^2+c^2) \] Since \[ a^2+b^2+c^2=1, \] we get \[ (a+b+c)^2\leq 3 \] So, \[ 1+2(ab+bc+ca)\leq 3 \] \[ 2(ab+bc+ca)\leq 2 \] \[ ab+bc+ca\leq 1 \] Hence, the maximum value is \[ 1 \]

Step 3: Find the minimum value.
Since \[ (a-b)^2+(b-c)^2+(c-a)^2\geq 0 \] Expanding, \[ 2(a^2+b^2+c^2)-2(ab+bc+ca)\geq 0 \] \[ a^2+b^2+c^2\geq ab+bc+ca \] This gives the upper bound. For lower bound, use \[ (a+b+c)^2\geq 0 \] So, \[ 1+2(ab+bc+ca)\geq 0 \] \[ 2(ab+bc+ca)\geq -1 \] \[ ab+bc+ca\geq -\frac{1}{2} \] Hence, the minimum value is \[ -\frac{1}{2} \]

Step 4: Write the set of extreme values.
The minimum value is \[ -\frac{1}{2} \] and the maximum value is \[ 1 \] Therefore, the set of extreme values is \[ \left\{-\frac{1}{2},1\right\} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\left\{-\frac{1}{2},1\right\}} \]