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Question

If (a+b+c)(ab+bc+ca)abc=x where a, b, c are distinct real numbers, then x is always greater than

A
30
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B
15
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C
17
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D
9
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Solution

The correct option is D 9
Given: (a+b+c)(ab+bc+ca)abc=x
We know that AM ≥ GM.
For any positive numbers a, b and c,
a+b+c3(abc)13 ........(i)
and for any positive numbers ab, bc and ca,
ab+bc+ca3(ab×bc×ca)13 ..........(ii)
ab+bc+ca3(abc)23
Multiplying (i) and (ii), we get,
(a+b+c)(ab+bc+ca)9(abc)13×(abc)23
(a+b+c)(ab+bc+ca)9abc
(a+b+c)(ab+bc+ca)9abc
(a+b+c)(ab+bc+ca)abc9
x9
Hence, the correct answer is option(4).

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