If $a^2+b^2+c^2=1, a, b, c \in R$, then the set of extreme values of $a b+b c+c a$ is

If $a^2+b^2+c^2=1, a, b, c \in R$, then the set of extreme values of $a b+b c+c a$ is
  1. $\left\{\frac{1}{2} ; 2\right\}$
  2. $\{-1,2\}$
  3. $\left\{-1, \frac{1}{2}\right\}$
  4. $\left\{\frac{-1}{2}, 1\right\}$

Solution

$\begin{aligned} & (a-b)^2+(b-c)^2+(c-a)^2 \geq 0 \\ \Rightarrow \quad & 2\left(a^2+b^2+c^2\right)-2(a b+b c+c a) \geq 0 \\ \Rightarrow \quad & a^2+b^2+c^2 \geq a b+b c+c a \\ \Rightarrow \quad & \quad 1 \geq a b+b c+c a \\ \text { Also } & (a+b+c)^2 \geq 0 \\ \Rightarrow & a^2+b^2+c^2+2(a b+b c+c a) \geq 0 \\ \Rightarrow & a b+b c+c a \geq-\frac{1}{2} \\ \therefore \quad & \quad-\frac{1}{2} \leq a b+b c+c a \leq 1\end{aligned}$

Asked in: AP EAMCET 2022 (07 Jul Shift 2)

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