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Question

If a,b,c are distinct positive real numbers such that a2+b2+c2=1, then value of ab+bc+ca, is:

A
less than 1
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B
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 A less than 1
We know 2(ab+ac+bc)=(a+b+c)2a2b2c2
2(ab+ac+bc)=(a+b+c)21
(ab+bc+ac)=(a+b+c)212
Now if a=b=c=13, then the above expression attains maximum value of 1.
Hence for distinct values of a,b and c,
(a+b+c)212<1
Hence, ab+bc+ac<1

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