$I f a^{2} + b^{2} + c^{2} = 2 t h e n t h e r a n g e o f a b + b c + c a i s$

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$[\frac{- 1}{2}, 1]$

$(1, \infty]$

$[\frac{- 1}{2}, \infty]$

$[- 1, 2]$

answer is C.

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Detailed Solution

$\begin{array}{l} G i v e n a^{2} + b^{2} + c^{2} = 2 \\ w e k n o w t h a t {(a + b + c)}^{2} \geq 0 \\ \Rightarrow a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) \geq 0 \\ \Rightarrow 2 + 2 (a b + b c + c a) \geq 0 \\ \Rightarrow a b + b c + c a \geq - 1 \\ A g a i n {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} \geq 0 \end{array}$

$\begin{array}{l} \Rightarrow 2 (a^{2} + b^{2} + c^{2}) - 2 (a b + b c + c a) \geq 0 \\ \Rightarrow a^{2} + b^{2} + c^{2} - (a b + b c + c a) \geq 0 \\ \Rightarrow 2 \geq a b + b c + c a \\ ∴ R a n g e = [- 1, 2] \end{array}$