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
B
Equal to 1
C
Greater than 1
D
Any real number

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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