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Question

If $$a, b$$ and $$c$$ are distinct positive real numbers and $$a^2+b^2+c^2=1$$, then $$ab+bc+ca$$ is-


A
Less than 1
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B
Equal to 1
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C
Greater than 1
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D
Any real number
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Solution

The correct option is B Less than $$1$$
Given $${ a }^{ 2 }{ +b }^{ 2 }+{ c }^{ 2 }=1$$
we know that, $${ (a }{ +b }+{ c) }^{ 2 }\ge 0$$
$$\Rightarrow { a }^{ 2 }{ +b }^{ 2 }+{ c }^{ 2 }+2{ (a }{ b+b }c+{ ca) }\ge 0$$
$$\Rightarrow 1+2{ (a }{ b+b }c+{ ca) }\ge 0$$
$$\Rightarrow a{ b+b }c+{ ca }\ge \frac { -1 }{ 2 } $$
Also, $$(b-c)^{ 2 }+(c-a)^{ 2 }+({ a-b })^{ 2 }\ge 0$$
$$\Rightarrow { 2(a }^{ 2 }{ +b }^{ 2 }+{ c }^{ 2 })-2(ab+bc+ca)\ge 0$$
$$\Rightarrow ab+bc+ca\le { a }^{ 2 }{ +b }^{ 2 }+{ c }^{ 2 }$$
$$\Rightarrow ab+bc+ca\le 1$$
Hence $$\frac { -1 }{ 2 } \le ab+bc+ca\le 1$$
Therefore, $$ab+bc+ca$$ is less than 1.

Mathematics

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