If $a, b a n d c$ are real numbers such that $a^{2} + b^{2} + c^{2} = 1$ . Then

$a b + b c + c a > \frac{1}{2}$

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$a b + b c + c a > - \frac{1}{2}$

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$a b + b c + c a > 1$

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$a b + b c + c a < - \frac{1}{2}$

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Solution

The correct option is B
$a b + b c + c a > - \frac{1}{2}$

Finding the value:
Given: $a, b a n d c a r e r e a l n u m b e r s .$
as we know ${(a + b + c)}^{2} \geq 0$
$\begin{array}{rcl} \Rightarrow & a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c \geq 0 \end{array}$
$\begin{array}{rcl} \Rightarrow & 1 + 2 (a b + b c + a c) \geq 0 [∵ a^{2} + b^{2} + c^{2} = 1] \\ \Rightarrow & 2 (a b + b c + a c) \geq - 1 \\ \Rightarrow & (a b + b c + a c) \geq \frac{- 1}{2} \end{array}$
Hence, option (B) is the correct answer.

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Similar questions

If a, b, c, d and p are different real numbers such that :

$(a^{2} + b^{2} + c^{2}) p^{2} - 2 (a b + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0$ , then show that a, b, c and d are in G.P.

\frac{b^{2} + c^{2}}{b + c} + \frac{c^{2} + a^{2}}{c + a} + \frac{a^{2} + b^{2}}{a + b} \geq a + b + c

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(a^{2} + b^{2} + c^{2}) x^{2} - 2 (a b + b c + c d) x + (b^{2} + c^{2} + d^{2}) \leq 0

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a^{2} + b^{2} + c^{2} = 1

a b + b c + c a > - \frac{1}{2}

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