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Question

If a,b,c,d be positive real numbers such that a+b+c+d=43, then the minimum value of 1b+c+d+1c+d+a+1d+a+b+1a+b+c is :

A
3
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B
6
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C
4
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D
8
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Solution

The correct option is C 4
Case of Minimum arise when a=b=c=d

So, a+b+c+d=4a=4b=4c=4d=43

So, a=b=c=d=13

Thus,

1a+b+c+1b+c+d+1c+d+a+1d+a+b=43/3=41=4

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