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If $a, b, c$ are distinct +ve real numbers and $a^{2}+b^{2}+c^{2}=1$ then $a b+b c+c a$ is

1) less than 1
2) equal to 1
3) greater than 1
4) any real no.
in Mathematics by Diamond (74,906 points) | 11 views

1 Answer

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Best answer





$\Rightarrow 2\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)>0$

$\Rightarrow 2>2(a b+b c+c a) ; \Rightarrow a b+b c+c a<1$

by Diamond (74,906 points)

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